Projectile Motion

Beginner

A projectile is anything thrown, kicked or launched that then moves under gravity alone: a basketball, a long jumper, the water from a garden hose. Its path is a curve called a parabola.

The trick is to split the motion in two. Sideways, nothing pushes or pulls the ball once it has left your hand (if we ignore the air), so it keeps the same horizontal speed. Up and down, gravity acts exactly as on a dropped ball: it adds 9.81 m/s of downward speed every second.

Throw a stone horizontally at 10 m/s from a cliff top. After 1 s it is 10 m out and has dropped about 4.9 m; after 2 s it is 20 m out and has dropped about 19.6 m.

Quick check: after 2 s, how fast is it moving forward, and how fast downward? Answer: still 10 m/s forward, and about 20 m/s downward.

Because the two motions ignore each other, a coin flicked off a table and a coin dropped from the same edge hit the floor together.

The launch angle matters. Thrown too flat, a ball lands quickly; thrown too steeply, it goes high but not far. On level ground and without air, 45° gives the longest throw, and 30° and 60° land at the same spot.

Newton pictured a cannon on a mountain firing ever faster: each ball lands farther away, until one falls exactly as fast as the round Earth curves away beneath it and never lands. That is an orbit.

In the animation, watch the grey dots, one every half second: sideways they are always the same distance apart, while vertically the gaps shrink on the way up and grow on the way down.

Intermediate

Axes, components and signs

A coin flicked off a table and a coin dropped from the same edge hit the floor together. Put the origin at the launch point, the $x$ axis horizontal in the direction of the throw and the $y$ axis vertically up. A ball launched at speed $v_0$ and angle $\theta$ above the horizontal has a velocity that is the hypotenuse of a right triangle whose sides are the components
$v_x=v_0\cos \theta$
$v_{y0}=v_0\sin \theta$

Both are signed, in metres per second: $v_y>0$ means rising, $v_y<0$ falling. Going back, the speed is $v=\sqrt{v_x^2+v_y^2}$. With up as positive, the vertical acceleration is $-g$, where $g=9.81m/{\mathrm{s}}^2$ is the size of the free-fall acceleration. Check how the angle is given: if it is measured from the vertical, sine and cosine swap places.

Two independent straight-line motions

Once the ball has left the hand, the only force on it, if we ignore the air, is its weight, straight down. Nothing pushes it sideways, so its horizontal velocity never changes: the horizontal motion is uniform. Vertically, the ball moves exactly like a ball thrown straight up at $v_0\sin \theta$, the case treated in Free Fall. The two motions share only the clock.

After a time $t$, therefore,
$x=v_0\cos \theta t$
$v_y=v_0\sin \theta -gt$
$y=v_0\sin \theta t-\frac{1}{2}g{t}^2$

A quick test of this independence: flick a coin off a table as you drop another from the same edge. Both start with $v_y=0$ and fall the same height, so they hit the floor together; the flicked one simply lands farther out.

Time of flight, height and range

On level ground, the time in the air, the height and the range all follow from the vertical motion plus one multiplication. The ball rises until $v_y=0$, at $t=v_0\sin \theta /g$, and the descent mirrors the climb, so it lands after twice that time. Putting the rise time into $y(t)$ gives the maximum height, and the horizontal speed times the time of flight gives the range:

• $T=\frac{2v_0\sin \theta }{g}$
• $H=\frac{(v_0\sin \theta {)}^2}{2g}$
• $R=\frac{v_0^2\sin 2\theta }{g}$

The time of flight and the height depend only on the vertical component $v_0\sin \theta$. The range needs both components, $R=v_0\cos \theta \cdot 2v_0\sin \theta /g$, and the identity $2\sin \theta \cos \theta =\sin 2\theta$ tidies it up.

Which angle throws farthest? A steep launch stays up long but moves slowly sideways; a flat one is fast sideways but lands almost at once. The range, their product, is largest in between: $\sin 2\theta$ reaches its maximum of 1 when $2\theta =90^{\circ }$, so on level ground $\theta =45^{\circ }$ is best, with $R_{\max }=v_0^2/g$. Angles equally far above and below 45° give the same range, because $\sin 2\theta =\sin (180^{\circ }-2\theta )$: 30° and 60° land at the same spot, and so do 40° and 50°.

Worked example: the throw in the animation

The ball leaves the ground at $v_0=20m/s$ and $\theta =50^{\circ }$. Its components are
$v_x=20\cos 50^{\circ }\approx 12.9m/s$
$v_{y0}=20\sin 50^{\circ }\approx 15.3m/s$

These are the values on the $v_x$ and $v_y$ readouts at launch. The ball reaches the top after $t=15.3/9.81\approx 1.56\mathrm{s}$ and lands after twice that, $T\approx 3.12\mathrm{s}$. Then
$H=\frac{15.32^2}{2\cdot 9.81}\approx 12.0\mathrm{m}$
$R=\frac{20^2\sin 100^{\circ }}{9.81}\approx 40.2\mathrm{m}$
and, as a check, $v_{x}T=12.9\cdot 3.12\approx 40.2\mathrm{m}$.

During the flight the gold $v_x$ readout stays at 12.9 m/s. The green $v_y$ readout drops by 9.81 m/s every second, passes through zero at 1.56 s, when the ball is at the top of the dashed line marked $H$, and reaches $-15.3m/s$ as the ball lands. The ball comes down 40.2 m away at 20 m/s, 50° below the horizontal: the descent is the climb played backwards.

At 40° the same speed reaches the same 40.2 m, but the ball rises only 8.4 m and lands after 2.62 s. At 45° it lands at 40.8 m, just 1.5 % farther than at 50°: near the best angle, the range is forgiving.

Worked example: a shot put

An athlete releases a shot 2.0 m above the ground at 13 m/s and 40° above the horizontal. How far away, measured horizontally, does it land? The level-ground formulas do not apply, because the shot lands 2.0 m below its starting height.

The components are $v_x=13\cos 40^{\circ }\approx 9.96m/s$ and $v_{y0}=13\sin 40^{\circ }\approx 8.36m/s$. The shot lands when $y=-2.0\mathrm{m}$, that is when $4.905t^2-8.36t-2.0=0$, where $4.905$ is $\frac{1}{2}g$. The positive root is
$t=\frac{8.36+\sqrt{8.36^2+4\cdot 4.905\cdot 2.0}}{2\cdot 4.905}\approx 1.92\mathrm{s}$
so the shot lands $x=9.96\cdot 1.92\approx 19.1\mathrm{m}$ away.

The level-ground formula would give $13^2\sin 80^{\circ }/9.81\approx 17.0\mathrm{m}$: the release height adds 2.1 m to the distance. The shot rises $8.36^2/(2\cdot 9.81)\approx 3.6\mathrm{m}$ above the hand and lands at 14.4 m/s, faster than it left, because it ends lower. When the landing point is below the launch, the best angle is a little under 45°, about 42° here; athletes release flatter still, because their arms push the shot faster at flatter angles.

Reading the strobe dots and the arrows

The grey dots mark the ball every 0.5 s. Sideways they are always $12.9\cdot 0.5\approx 6.4\mathrm{m}$ apart: equal distances in equal times, the signature of uniform motion. Vertically, counting from the launch point, the ball climbs 6.4, 4.0 and 1.5 m from one half-second mark to the next, then drops 0.9, 3.4 and 5.8 m. Each step is 2.45 m less upward, or more downward, than the one before, because $g(0.5\mathrm{s}{)}^2=2.45\mathrm{m}$: the pattern of a dropped ball, started with an upward push.

The arrows tell the same story. The blue $v$ arrow is always tangent to the dashed trail. The gold $v_x$ arrow keeps its length from launch to landing, while the green $v_y$ arrow shrinks, vanishes at the top and then grows pointing down. So the blue arrow is shortest at the top, where the speed equals $v_x$, 12.9 m/s, and it is as long at landing as at launch.

Common mistakes

• The velocity is not zero at the top. Only $v_y$ is; the ball still moves sideways at 12.9 m/s, its lowest speed of the whole flight.
• Nor is the acceleration zero at the top. It is still 9.81 m/s² downward, so 0.1 s after the top $v_y$ is already $-0.98m/s$.
• A faster sideways throw does not stay up longer. Rolled off a table 1.0 m high at 2 m/s or at 4 m/s, a ball lands after 0.45 s either way, 0.9 m or 1.8 m out.
• The range needs $\sin 2\theta$, not $\sin \theta$. With $\sin 50^{\circ }$ instead of $\sin 100^{\circ }$, the animation's throw would seem to land at 31.2 m, not 40.2 m.
• The range formula assumes landing at the launch height. For the shot put it gives 17.0 m instead of 19.1 m.

Where you meet it

Ball games, the long jump, the jet of a hose or a fountain (each drop follows its own parabola), springboard dives and fireworks all follow these equations while the air matters little. That holds well for a heavy shot, less well for a football and not at all for a shuttlecock, a case for Drag and Terminal Velocity.

Galileo showed in 1638 that a projectile's path is a parabola, by combining a uniform horizontal motion with a uniformly accelerated fall. Newton's Laws of Motion explain why the two motions are independent. Followed far enough, the same reasoning leads to orbits: launched sideways at about 7.9 km/s, a body falls around the Earth as fast as the ground curves away, the subject of Gravitation and Orbits.

Advanced

Newton's law, one component at a time

Everything in this lesson rests on one move: splitting a motion in a plane into two motions along straight lines. Put the launch point at the origin, with $x$ horizontal and $y$ up. The ball leaves with the speed $v_0$ at the angle $\theta$ above the horizontal, so its launch velocity ${\vec{v}}_0$ has the components $v_0\cos \theta$ and $v_0\sin \theta$. Write $g=9.81m/{\mathrm{s}}^2$ for the size of the free-fall acceleration, so that $\vec{g}=(0,-g)$ points straight down.

Once the ball is in flight, and as long as air resistance is neglected, the only force on it is its weight $m\vec{g}$. Newton's second law reads $m\vec{a}=m\vec{g}$. Two vectors are equal only when their components are equal, so this one equation splits into two, where a dot means a derivative with respect to time:
$m\ddot{x}=0,m\ddot{y}=-mg$

Two things follow at once. The equations do not mix: $y$ never appears in the first and $x$ never appears in the second, so each is a straight-line problem of its own. And the mass cancels, so a marble and a cannonball launched with the same velocity follow the same path. Sideways the ball moves at a constant velocity; vertically it is in free fall.

Integrate each equation once to get the velocity, then again to get the position. Each integration brings in a constant, fixed by the state at $t=0$: the ball starts at the origin with the velocity components above. The result is
$\dot{x}=v_0\cos \theta ,x=v_0\cos \theta t$
$\dot{y}=v_0\sin \theta -gt,y=v_0\sin \theta t-\frac{1}{2}g{t}^2$

Four constants were needed: two starting coordinates and two starting velocity components. Motion in a plane always needs four starting numbers, twice as many as motion along a line.

The numbers below refer to the animation's throw: $v_0=20m/s$ at $\theta =50^{\circ }$. Its horizontal component is $v_x=20\cos 50^{\circ }\approx 12.9m/s$ and its initial vertical component $20\sin 50^{\circ }\approx 15.3m/s$.

The position as a vector sum

The two component results can be packed back into a single vector equation. It is worth doing, because the vector form shows at a glance how gravity pulls the ball away from the line it was aimed along.

Legend

$\theta$	Angle of the line of aim above the horizontal.	$\mathrm{rad}$
${\vec{v}}_{0}t$	Where the ball would be without gravity: the launch velocity times the time, along the line of aim.	$\mathrm{m}$
$\frac{1}{2}\vec{g}{t}^2$	Drop caused by gravity: straight down, growing as $t^2$.	$\mathrm{m}$
$\vec{r}$	Position of the ball: the arrow from the launch point to the ball.	$\mathrm{m}$
$M$	Point on the line of aim where a monkey hangs at the instant of launch.

Figure 1 draws the equation
$\vec{r}(t)={\vec{v}}_{0}t+\frac{1}{2}\vec{g}{t}^2$
Read it as two arrows placed tip to tail. The arrow ${\vec{v}}_{0}t$ starts at the launch point and runs along the line of aim, at the angle $\theta$ above the horizontal. It ends where the ball would be at the time $t$ if there were no gravity. From its tip the arrow $\frac{1}{2}\vec{g}{t}^2$ points straight down: it is how far the ball has fallen below the line of aim. The position vector $\vec{r}$, from the launch point to the ball, is the sum of the two, and the red curve is the path so far.

The first arrow grows in proportion to $t$, the second in proportion to $t^2$. Early in the flight the drop is tiny compared with the distance along the line of aim; later it takes over, so the ball sinks ever faster below that line.

This settles a classic puzzle. A monkey hangs from a branch at the point $M$ on the line of aim, and lets go at the very instant the ball is fired straight at it. Does the ball pass over the falling monkey? No. Figure 1 shows the moment when ${\vec{v}}_{0}t$ has reached $M$. The ball is then $\frac{1}{2}g{t}^2$ below the line of aim, directly beneath $M$. The monkey, falling from rest for the same time $t$, has dropped by exactly the same distance. So the ball meets the monkey whatever the launch speed, provided it gets there before both reach the ground: gravity lowers both by the same amount, so aiming straight at the monkey is exactly right.

The velocity has the same structure, $\vec{v}(t)={\vec{v}}_0+\vec{g}t$. Its horizontal part never changes, and its vertical part loses 9.81 m/s every second. Draw the velocity arrows of successive instants from one common point: their tips march straight down a vertical line, 4.9 m/s for every half second.

Eliminating time: the parabola

The equations so far say where the ball is at each instant. To see the shape of the path itself, remove the time.

Because $x$ grows at the constant rate $v_0\cos \theta$, the ball reaches a given $x$ at the time $t=x/(v_0\cos \theta )$. Substituting this into $y(t)$ gives the height as a function of the horizontal distance:
$y=x\tan \theta -\frac{g{x}^2}{2v_0^2{\cos }^2\theta }$

Read it term by term. The first term, $x\tan \theta$, is the line of aim. The second is the drop $\frac{1}{2}g{t}^2$ of Figure 1, now written in terms of $x$. Because the drop grows as $x^2$, the path is a parabola that opens downwards.

Two landmarks follow. The ball is back at launch height where $y=0$: at $x=0$, and at the range
$R=\frac{v_0^2\sin 2\theta }{g}$
where the identity $2\sin \theta \cos \theta =\sin 2\theta$ has been used. The top of the arc is where the path is level. The slope is $dy/dx=\tan \theta -gx/(v_0^2{\cos }^2\theta )$, which vanishes at $x=v_0^2\sin \theta \cos \theta /g=R/2$. So the apex lies exactly halfway along, as the symmetry of a parabola requires, at the height $H=v_0^2{\sin }^2\theta /(2g)$.

For the animation's throw the path is $y=1.192x-0.02968x^2$, with $x$ and $y$ in metres. It returns to the ground at 40.2 m and peaks at 12.0 m, when $x=20.1\mathrm{m}$.

Aiming at a target: two angles and the edge of reach

So far the launch was given and we asked where the ball goes. A thrower faces the reverse question: with a fixed speed, at which angle must the ball leave to hit a given target? The same trajectory equation answers it.

Legend

$v_0$	Launch speed, the same for both throws and for the dashed parabola.	$m/s$
${\theta }_1$	Launch angle of the flat throw.	$\mathrm{rad}$
${\theta }_2$	Launch angle of the lofted throw.	$\mathrm{rad}$
$T$	The target.
$x$	Horizontal distance from the launch point to the target.	$\mathrm{m}$
$y$	Height of the target above the launch point.	$\mathrm{m}$
$v_0^2/(2g)$	Greatest height reachable at speed $v_0$, by a throw straight up: the top of the dashed parabola.	$\mathrm{m}$
$v_0^2/g$	Greatest range on level ground, reached at $45^{\circ }$: where the dashed parabola meets the ground.	$\mathrm{m}$

In Figure 2 the ball leaves the launch point with the speed $v_0$ and must hit the target $T$, a horizontal distance $x$ away and a height $y$ above the launch point. Two paths do it, a flat throw at the angle ${\theta }_1$ and a lofted throw at ${\theta }_2$, and they cross at $T$.

To find these angles, treat $\theta$ as the unknown. The trajectory equation contains both $\tan \theta$ and $1/{\cos }^2\theta$. The identity $1/{\cos }^2\theta =1+{\tan }^2\theta$ turns it into an equation for $u=\tan \theta$ alone:
$y=xu-\frac{g{x}^2}{2v_0^2}\left(1+u^2\right)$
The quantity $D=g{x}^2/(2v_0^2)$ is a length: the drop of a ball launched horizontally at $v_0$ over the distance $x$. With it, collecting the powers of $u$ gives a quadratic, the aiming equation:
$D{u}^2-xu+(y+D)=0$

A quadratic has two, one or no real roots, and each case has a meaning. Two roots: a flat and a lofted shot both hit the target, the two paths of Figure 2. A double root: the target lies exactly on the limit of reach, and only one angle gets there. No real root: the target is out of range at this speed.

The product of the two roots leads to a simple result. For any quadratic it equals the constant term divided by the coefficient of $u^2$, here $1+y/D$. On level ground, $y=0$, the product is 1, so $\tan {\theta }_1\tan {\theta }_2=1$ and ${\theta }_2=90^{\circ }-{\theta }_1$: the two angles are complementary, and throws at 30° and at 60° land at the same spot. For a target above the launch point, as in Figure 2, the product is larger than 1. Then $\tan ({\theta }_1+{\theta }_2)=(\tan {\theta }_1+\tan {\theta }_2)/(1-\tan {\theta }_1\tan {\theta }_2)$ is negative, so the two angles add up to more than $90^{\circ }$.

The double root traces the limit of reach itself. It occurs when the discriminant vanishes, $x^2=4D(y+D)$. Solving for $y$, and putting back $D=g{x}^2/(2v_0^2)$, gives the edge of reach:
$y=\frac{v_0^2}{2g}-\frac{g{x}^2}{2v_0^2}$

This curve, often called the safety parabola, is the dashed curve of Figure 2. Every path launched at the speed $v_0$ touches it at one point and stays inside it, so it encloses every point the ball can reach. It stands $v_0^2/(2g)$ above the launch point, the height of a throw straight up, and meets the ground at $v_0^2/g$. That is the largest range on level ground, reached at 45°, where $\sin 2\theta =1$. At 20 m/s these lengths are 20.4 m and 40.8 m. Paths launched above 45° touch the curve during the flight, as the lofted path of Figure 2 does just past its own apex; flatter paths would touch it only if continued below the ground.

The velocity graph, and speed against height

Two more tools describe the flight without solving anything new: the graph of the vertical velocity, and a relation between speed and height that does not involve the time at all.

The expert layer of the animation plots $v_y(t)$. Because ${\dot{v}}_y=-g$ is constant, the graph is a straight line of slope $-g=-9.81m/{\mathrm{s}}^2$, here from 15.3 m/s down to $-15.3m/s$. It crosses zero at $t=15.3/9.81\approx 1.56\mathrm{s}$, the moment of the apex. The area under a velocity graph is the displacement, so the triangle above the axis, $\frac{1}{2}\cdot 1.562\cdot 15.32\approx 12.0\mathrm{m}$, is the height $H$. The equal triangle below the axis is the way back down, and the ball is at $y=0$ again at 3.12 s.

Speed and height are linked without time. Start from ${\dot{v}}_y=-g$ and multiply both sides by $v_y=\dot{y}$. The left side becomes $\frac{1}{2}d(v_y^2)/dt$ and the right side $-gdy/dt$. Both sides are now time derivatives, so integrating from the launch gives $v_y^2=v_0^2{\sin }^2\theta -2gy$. The horizontal component never changes, so adding $v_x^2=v_0^2{\cos }^2\theta$ to both sides gives
$v^2=v_0^2-2gy$

The speed depends only on the height, not on the time or on the direction of motion. The ball passes each level at the same speed going up and coming down, and it lands at 20 m/s, its launch speed. At the apex $v^2=400-2\cdot 9.81\cdot 12.0\approx 165{\mathrm{m}}^2/{\mathrm{s}}^2$, which is just $v_x^2$, as it must be when the ball moves horizontally. Multiplied by $m/2$, the relation reads $\frac{1}{2}m{v}^2+mgy=\frac{1}{2}m{v}_0^2$: kinetic plus potential energy stays constant.

Curvature of the path

A parabola is not a circle, but at each point one circle fits it best, and the radius of that circle measures how sharply the path bends there. Finding it shows the two separate jobs gravity does: it turns the ball and it changes its speed.

Legend

$P$	Point of the path where the velocity makes the angle $\alpha$ with the horizontal.
$\vec{v}$	Velocity at $P$, along the path.	$m/s$
$\alpha$	Angle between $\vec{v}$ and the horizontal; the same angle lies between $\vec{g}$ and the direction across the path.	$\mathrm{rad}$
$\vec{g}$	Acceleration of gravity, straight down.	$m/{\mathrm{s}}^2$
$g\sin \alpha$	Part of $\vec{g}$ along the path: it points against $\vec{v}$ while the ball rises, and changes only the speed.	$m/{\mathrm{s}}^2$
$g\cos \alpha$	Part of $\vec{g}$ across the path, towards $C$: it changes only the direction.	$m/{\mathrm{s}}^2$
$\rho$	Radius of curvature at $P$: the distance from $P$ to $C$.	$\mathrm{m}$
$C$	Centre of the circle that fits the path at $P$.

In Figure 3 the ball is at the point $P$, moving with the velocity $\vec{v}$ along the path, at the angle $\alpha$ above the horizontal. Split $\vec{g}$ into a part along the path and a part across it, at right angles to $\vec{v}$; the small square at $P$ marks that right angle. The along and across directions are just the horizontal and the vertical turned through $\alpha$. Since $\vec{g}$ points down the vertical, it makes the same angle $\alpha$ with the across direction, as the arc between $\vec{g}$ and $g\cos \alpha$ shows. So the part along the path is $g\sin \alpha$, pointing back against $\vec{v}$ while the ball rises, and the part across it is $g\cos \alpha$.

The two parts do different jobs. An acceleration along the velocity changes only the speed; one across the velocity changes only the direction. So only $g\cos \alpha$ bends the path. It points to the centre $C$ of the circle that fits the path at $P$, the distance $\rho$ away. A body moving at the speed $v$ on a circle of radius $\rho$ needs the acceleration $v^2/\rho$ towards the centre, so $g\cos \alpha =v^2/\rho$. With $\cos \alpha =v_x/v$, the radius of curvature is
$\rho =\frac{v^2}{g\cos \alpha }=\frac{v^3}{g{v}_x}$

Because $v_x$ never changes, $\rho$ is smallest where the speed is smallest: at the apex. There $\alpha =0$, gravity lies wholly across the path, and $\rho =v_x^2/g=12.9^2/9.81\approx 17\mathrm{m}$. At launch $\rho =20^3/(9.81\cdot 12.9)\approx 63\mathrm{m}$. The arc is sharpest at the top.

The other part, $g\sin \alpha$, changes the speed. While the ball rises it points against $\vec{v}$ and slows the ball. On the way down $\vec{v}$ points below the horizontal, the part along the path points the same way as $\vec{v}$, and the ball speeds up.

Air resistance and numerical stepping

Real balls fly through air. Drag spoils the neat formulas above, but a computer can still follow the motion step by step, and the method works for any force.

Legend

$R$	Range with air resistance: the distance from the launch point to the landing point.	$\mathrm{m}$
$R_0$	Range without air: where the dashed path lands.	$\mathrm{m}$
$\vec{v}$	Velocity of the ball, drawn at launch and at landing.	$m/s$
${\vec{a}}_d$	Acceleration caused by drag: along the path, against $\vec{v}$, of size $g(v/v_t{)}^2$.	$m/{\mathrm{s}}^2$

Figure 4 sets the animation's throw in air (red) beside the same throw without air (dashed). The drag force always opposes the velocity, and for a ball at everyday speeds it grows roughly as the square of the speed. Call $v_t$ the terminal speed: the speed at which drag equals the weight, reached by a ball dropped from a great height. The drag acceleration then has the size $g(v/v_t{)}^2$ and points against $\vec{v}$; in vector form it is $-kv\vec{v}$, with $k=g/v_t^2$. In Figure 4 each arrow ${\vec{a}}_d$ points back along the path, against $\vec{v}$, and the one at landing is the shorter, because the ball lands slower than it left.

Adding gravity, the equations of motion become
$\ddot{x}=-kv\dot{x},\ddot{y}=-g-kv\dot{y}$
with $v=\sqrt{{\dot{x}}^2+{\dot{y}}^2}$. The speed contains both velocity components, so each equation now depends on the other's motion: the horizontal and vertical motions are coupled, and no formula gives the path.

It can still be followed step by step. Choose a small time step $\epsilon$. From the present position and velocity compute the acceleration $\vec{a}$, then move forward by one step,
$\vec{v}\to \vec{v}+\vec{a}\epsilon ,\vec{r}\to \vec{r}+\vec{v}\epsilon$
and repeat until the ball lands. Over one short step the acceleration and the velocity hardly change, so treating them as constant makes only a small error. In this simplest scheme, the Euler method, the total error shrinks in proportion to $\epsilon$: halving the step halves the error. Steps of 0.01 s already give the range below to within about 0.2 m. Nothing in the method depends on the form of the force, so it works for any force law.

The red path of Figure 4 is such a calculation, for a ball whose terminal speed is 30 m/s. The drag starts at $(20/30{)}^2\approx 0.44$ of the weight and is down to $(15.6/30{)}^2\approx 0.27$ of it at landing. The range $R$ falls from $R_0=40.2\mathrm{m}$ to 30.0 m, and the greatest height from 12.0 m to 10.1 m.

The arc is no longer symmetric. On the way up drag adds to gravity in slowing the climb; on the way down it opposes the fall. So the climb takes 1.39 s and the descent 1.48 s. Drag also keeps reducing the horizontal speed, so the ball covers more ground while climbing than while falling: the apex sits 15.9 m out, past the middle. The ball lands at 15.6 m/s, 57° below the horizontal, steeper than its 50° launch.

Drag also shifts the best angle. The angle that gives the longest range drops to about 43°, because a higher throw stays up longer and so loses more to drag. The faster the launch compared with $v_t$, the lower the best angle.

Worked problem: two ways to hit a target

Apply the aiming equation to the target of Figure 2, with the animation's launch speed of 20 m/s. At what angles must you throw to hit a target 30 m away and 5.0 m above the launch point?

First the drop: $D=g{x}^2/(2v_0^2)=9.81\cdot 900/800\approx 11.04\mathrm{m}$, so $y+D\approx 16.04\mathrm{m}$. With $x=30\mathrm{m}$ the aiming equation reads
$11.04u^2-30u+16.04=0$
and the quadratic formula gives
$u=\frac{30\pm \sqrt{900-4\cdot 11.04\cdot 16.04}}{2\cdot 11.04}=\frac{30\pm 13.84}{22.08}$
so $u\approx 0.732$ or $u\approx 1.986$, that is $\theta \approx 36.2^{\circ }$ or $\theta \approx 63.3^{\circ }$. These are the angles ${\theta }_1$ and ${\theta }_2$ of Figure 2. They add up to 99.5°, more than $90^{\circ }$, as expected for a target above the launch point.

The flight times follow from $t=x/(v_0\cos \theta )$: $30/(20\cos 36.2^{\circ })\approx 1.86\mathrm{s}$ for the flat shot and $30/(20\cos 63.3^{\circ })\approx 3.34\mathrm{s}$ for the lofted one. Both arrive at the same speed, $\sqrt{400-2\cdot 9.81\cdot 5.0}\approx 17.4m/s$, because the speed depends only on the height.

They arrive from different directions, though. The flat shot has already passed its 7.1 m apex and comes in 22° below the horizontal; the lofted one drops in at 59° from an apex 16.3 m up. This is why basketball players favour steep arcs: to a ball dropping in steeply, the hoop presents a wider opening.

Could the same throw reach a target 10 m up at that distance? The edge of reach at $x=30\mathrm{m}$ lies at $20.39-11.04\approx 9.35\mathrm{m}$. The target is above it, the aiming equation has no real root, and no angle works. A target exactly 9.35 m up is hit by a single shot. The double root is $u=x/(2D)$, which here is $\tan \theta =v_0^2/(gx)\approx 1.36$, that is $\theta \approx 53.7^{\circ }$: the flat and the lofted solutions have merged into one.

Limits of validity

Everything above rests on three idealisations: a ball that behaves as a point, flat ground and a uniform pull of gravity. Each holds well for thrown balls, and each fails in a known way.

The model treats the ball as a point with no spin. A spinning ball also feels a lift force across its velocity, the Magnus effect, which curls a free kick and keeps a backspinning golf ball aloft. Drag matters when the speed is a sizeable fraction of the terminal speed, since drag divided by weight is $(v/v_t{)}^2$. A 7.26 kg shot has a terminal speed of about 150 m/s, so at 13 m/s its drag is under 1 % of its weight and the parabola is excellent.

Flat ground and a uniform $\vec{g}$ are approximations too. With the Earth's radius $R_E\approx 6371\mathrm{km}$, the ground curves down over the animation's 40 m by only $x^2/(2R_E)\approx 0.1\mathrm{mm}$. Gravity weakens as the inverse square of the distance from the Earth's centre, so at the 12 m apex $g$ is smaller by the fraction $2H/R_E\approx 4\times 10^{-6}$.

These effects grow with the range. A launch at 1 km/s and 45°, with a vacuum range of 102 km, would come down where the ground has dropped by 0.8 km, though air would dominate long before that. The Earth's rotation shifts the landing point of the 3.1 s throw by under a centimetre, but that of a shell flying for a minute by tens of metres.

Strictly, gravity points to the Earth's centre rather than in one fixed direction, and the trajectory is an arc of an ellipse with one focus at that centre. Over a short stretch, where the directions to the centre are almost parallel, the ellipse cannot be told apart from the parabola. Launched sideways at $\sqrt{g{R}_E}\approx 7.9km/s$, and ignoring the air, a body falls around the Earth exactly as fast as the ground curves away beneath it: the projectile has become a satellite, and the thrown ball and the Moon obey one law.

Where this leads

Newton's Laws of Motion supply the component equations. The relation $v^2=v_0^2-2gy$ is energy conservation, developed in Work, Energy and Power. At the apex gravity plays the part of the centripetal acceleration of Circular Motion, $g=v_x^2/\rho$. The coupled drag equations are the subject of Drag and Terminal Velocity, and Newton's cannon and the ellipse belong to Gravitation and Orbits.

A ball dropped from the window of a steadily moving train falls straight down for a passenger but traces a parabola for someone on the platform: one motion seen from two frames, as in Vectors and Relative Motion. And a charged particle in the uniform field between two charged plates follows the same parabola, with $g$ replaced by $qE/m$; oscilloscope tubes and some ink-jet printers steer electrons and ink droplets this way (Electric Field and Potential).