Projectile motion: equations, maximum height, time and range

A projectile is an object given an initial velocity and then allowed to move under gravity. In the standard A level model, gravity is the only force after projection. The horizontal velocity is therefore constant, while the vertical velocity changes at the constant rate $-g$.

The central idea is simple but essential: horizontal and vertical motion are solved separately, but they share the same time $t$.

Prerequisites

You should be able to:

• resolve a vector using sine and cosine;
• use constant acceleration equations with a consistent sign convention;
• interpret displacement and velocity as vectors using two dimensional motion;
• solve linear and quadratic equations;
• use exact trigonometric values and inverse trigonometric functions.

Unless a question states otherwise, use $g=9.8\ \mathrm{m\,s^{-2}}$ and give non-exact answers to a sensible degree of accuracy.

The projectile model

The usual model assumes that:

• the projectile is a particle;
• air resistance is ignored;
• gravity is uniform;
• the ground is fixed and horizontal unless stated otherwise.

Choose $x$ horizontally and $y$ vertically upwards. Once the projectile is released,

$$a_x=0, \qquad a_y=-g.$$

If it is projected with speed $U$ at angle $\theta$ above the horizontal, its initial velocity components are

$$u_x=U\cos\theta, \qquad u_y=U\sin\theta.$$

Hence, taking the point of projection as $(0,0)$,

$$\boxed{x=U\cos\theta\,t}$$

and

$$\boxed{y=U\sin\theta\,t-\frac12gt^2}.$$

The velocity at time $t$ is

$$\boxed{v_x=U\cos\theta, \qquad v_y=U\sin\theta-gt}.$$

These equations contain the whole topic.

One time, two directions

Do not find a separate horizontal time and vertical time. At a particular instant, the same $t$ belongs in both component equations. Usually one direction provides $t$, then the other direction provides the requested quantity.

A reliable method

1. Draw a sketch and mark the launch point, positive directions and target level.
2. Resolve the initial velocity into horizontal and vertical components.
3. Write separate data lists for the two directions.
4. Use the component containing enough known information to find $t$.
5. Substitute that same $t$ into the other component.
6. Reject times outside the physical interval, such as a negative time or a time after impact.
7. Check units and whether the size and direction of the answer are reasonable.

The horizontal component has $a_x=0$, so it rarely needs a full SUVAT formula. Usually $x=u_xt$ is enough.

Position and velocity at a given time

Worked example 1: finding position, speed and direction

A particle is projected from the origin with speed $25\ \mathrm{m\,s^{-1}}$ at $30^\circ$ above the horizontal. Find its position and velocity after $1.5$ seconds. Hence find its speed and direction of motion. Use $g=9.8\ \mathrm{m\,s^{-2}}$.

Resolve the initial velocity:

$$u_x=25\cos30^\circ=\frac{25\sqrt3}{2}, \qquad u_y=25\sin30^\circ=12.5.$$

Horizontal position:

$$x=\frac{25\sqrt3}{2}(1.5) =\frac{75\sqrt3}{4} \approx32.5\ \mathrm m.$$

Vertical position:

$$\begin{aligned} y&=12.5(1.5)-\frac12(9.8)(1.5)^2\\ &=7.725\ \mathrm m. \end{aligned}$$

Thus the position is approximately

$$\boxed{(32.5,\ 7.73)\ \mathrm m}.$$

The horizontal velocity remains constant:

$$v_x=\frac{25\sqrt3}{2}\approx21.7\ \mathrm{m\,s^{-1}}.$$

Vertically,

$$v_y=12.5-9.8(1.5)=-2.2\ \mathrm{m\,s^{-1}}.$$

The negative vertical component shows that the particle is descending. Its speed is the magnitude of its velocity:

$$|\mathbf v|=\sqrt{v_x^2+v_y^2} =\sqrt{\left(\frac{25\sqrt3}{2}\right)^2+(-2.2)^2} \approx21.8\ \mathrm{m\,s^{-1}}.$$

If $\alpha$ is the angle below the horizontal,

$$\tan\alpha=\frac{|v_y|}{v_x} =\frac{2.2}{25\sqrt3/2},$$

so $\alpha\approx5.80^\circ$. The particle is moving at

$$\boxed{21.8\ \mathrm{m\,s^{-1}}\text{ at }5.80^\circ\text{ below the horizontal}}.$$

Velocity direction is not the launch angle

Gravity changes $v_y$ but not $v_x$. Except at launch, the direction of motion must be found from the current velocity components, not from $\theta$.

Self-check 1

A ball is projected at $20\ \mathrm{m\,s^{-1}}$ at $40^\circ$ above the horizontal. Find its velocity vector after $1$ second and its speed then.

Answer

$$\mathbf v =(20\cos40^\circ)\mathbf i+(20\sin40^\circ-9.8)\mathbf j.$$

Therefore

$$\boxed{\mathbf v=(15.3\mathbf i+3.06\mathbf j)\ \mathrm{m\,s^{-1}}}$$

to $3$ significant figures, and

$$\boxed{|\mathbf v|=15.6\ \mathrm{m\,s^{-1}}}.$$

Maximum height

At the highest point, the vertical velocity is zero:

$$v_y=0.$$

The whole velocity is not zero because the horizontal component usually remains $U\cos\theta$.

From $v_y=U\sin\theta-gt$, the time to maximum height is

$$\boxed{t_{\max}=\frac{U\sin\theta}{g}}.$$

Using $v_y^2=u_y^2+2a_ys_y$ gives the maximum vertical displacement above the launch point:

$$0=(U\sin\theta)^2-2gH,$$

so

$$\boxed{H=\frac{U^2\sin^2\theta}{2g}}.$$

Worked example 2: highest point

A stone is projected at $28\ \mathrm{m\,s^{-1}}$ at $45^\circ$ above the horizontal. Find the time at which it reaches its highest point and the coordinates of that point.

The initial components are

$$u_x=u_y=28\left(\frac{1}{\sqrt2}\right)=14\sqrt2.$$

At maximum height, $v_y=0$:

$$0=14\sqrt2-9.8t,$$

so

$$t=\frac{14\sqrt2}{9.8}=\frac{10\sqrt2}{7} \approx2.02\ \mathrm s.$$

The horizontal coordinate is

$$x=14\sqrt2\left(\frac{10\sqrt2}{7}\right)=40\ \mathrm m.$$

For the height, use the vertical equation without time:

$$0=(14\sqrt2)^2-2(9.8)y.$$

Therefore

$$y=20\ \mathrm m.$$

The highest point is

$$\boxed{(40,20)\ \mathrm m},$$

reached after $\boxed{2.02\ \mathrm s}$.

Self-check 2

A particle is projected at $30\ \mathrm{m\,s^{-1}}$ at an angle $\theta$ where $\sin\theta=3/5$. Find its maximum height above the launch point.

Answer

The initial vertical velocity is

$$u_y=30\left(\frac35\right)=18\ \mathrm{m\,s^{-1}}.$$

Hence

$$H=\frac{18^2}{2(9.8)} =\boxed{16.5\ \mathrm m}$$

to $3$ significant figures.

Returning to the launch level

If the projectile lands at the same vertical level from which it was launched, set $y=0$:

$$0=U\sin\theta\,t-\frac12gt^2.$$

Factorise before dividing by $t$:

$$t\left(U\sin\theta-\frac12gt\right)=0.$$

The root $t=0$ represents launch. The later root is the time of flight:

$$\boxed{T=\frac{2U\sin\theta}{g}}.$$

The horizontal range is

$$R=U\cos\theta\,T =\frac{2U^2\sin\theta\cos\theta}{g}.$$

Since $2\sin\theta\cos\theta=\sin2\theta$,

$$\boxed{R=\frac{U^2\sin2\theta}{g}}.$$

These formulae apply only when landing and launch are at the same level.

Worked example 3: time of flight and range

A golf ball is projected from level ground at $35\ \mathrm{m\,s^{-1}}$ at $38^\circ$ above the horizontal. Ignore air resistance. Find its time of flight and horizontal range.

The time of flight is

$$\begin{aligned} T&=\frac{2(35)\sin38^\circ}{9.8}\\ &\approx4.40\ \mathrm s. \end{aligned}$$

Then

$$\begin{aligned} R&=35\cos38^\circ(4.40)\\ &\approx121\ \mathrm m. \end{aligned}$$

Equivalently,

$$R=\frac{35^2\sin76^\circ}{9.8}\approx121\ \mathrm m.$$

Thus

$$\boxed{T=4.40\ \mathrm s, \qquad R=121\ \mathrm m}.$$

Complementary launch angles

For fixed $U$, equal launch and landing levels, and angles between $0^\circ$ and $90^\circ$,

$$R=\frac{U^2\sin2\theta}{g}.$$

The angles $\theta$ and $90^\circ-\theta$ give the same range because

$$\sin(2(90^\circ-\theta)) =\sin(180^\circ-2\theta) =\sin2\theta.$$

One path is low and fast, while the other is high and slow. The maximum range occurs when $\sin2\theta=1$, so

$$\boxed{\theta=45^\circ}.$$

This conclusion depends on the model and equal launch and landing heights. It need not hold from a cliff or when air resistance matters.

Self-check 3

A projectile is launched from level ground at $24\ \mathrm{m\,s^{-1}}$. Find its greatest possible range and the launch angle that produces it.

Answer

The greatest range occurs at $45^\circ$:

$$R_{\max}=\frac{U^2}{g} =\frac{24^2}{9.8} =\boxed{58.8\ \mathrm m}$$

to $3$ significant figures.

The Cartesian equation of the trajectory

The component equations use time as a parameter:

$$x=U\cos\theta\,t, \qquad y=U\sin\theta\,t-\frac12gt^2.$$

To obtain $y$ directly in terms of $x$, rearrange the horizontal equation:

$$t=\frac{x}{U\cos\theta}.$$

Substitute into the vertical equation:

$$\begin{aligned} y &=U\sin\theta\left(\frac{x}{U\cos\theta}\right) -\frac12g\left(\frac{x}{U\cos\theta}\right)^2\\ &=x\tan\theta-\frac{gx^2}{2U^2\cos^2\theta}. \end{aligned}$$

Therefore

$$\boxed{y=x\tan\theta-\frac{gx^2}{2U^2\cos^2\theta}}.$$

This is a quadratic in $x$ with a negative $x^2$ coefficient, so the ideal trajectory is a downward opening parabola.

Worked example 4: height at a horizontal distance

A particle is projected from the origin at $20\ \mathrm{m\,s^{-1}}$ at $30^\circ$ above the horizontal. Find its height when it is $15$ m horizontally from the launch point.

Using the trajectory equation,

$$y=15\tan30^\circ -\frac{9.8(15)^2}{2(20)^2\cos^230^\circ}.$$

Since $\tan30^\circ=1/\sqrt3$ and $\cos^230^\circ=3/4$,

$$\begin{aligned} y &=5\sqrt3-\frac{2205}{600}\\ &\approx4.99\ \mathrm m. \end{aligned}$$

So the particle is

$$\boxed{4.99\ \mathrm m}$$

above the launch level.

Two valid times at one height

A horizontal line below the maximum may meet the trajectory twice. Algebraically, the vertical equation then has two positive roots: one while the projectile rises and one while it falls. Use horizontal position or the wording of the question to choose the required event.

Launching from above the landing level

When a projectile starts at height $h$ above horizontal ground, either take the launch point as $y=0$ and the ground as $y=-h$, or take ground as $y=0$ and write

$$y=h+U\sin\theta\,t-\frac12gt^2.$$

Both approaches are correct. Do not mix their displacement conventions.

Worked example 5: horizontal projection from a cliff

A stone is projected horizontally at $12\ \mathrm{m\,s^{-1}}$ from a cliff $19.6$ m above level ground. Find the time to hit the ground, its horizontal distance from the foot of the cliff, and its speed on impact.

Take the launch point as the origin, with upwards positive. The initial components are

$$u_x=12, \qquad u_y=0.$$

At impact, the vertical displacement is $y=-19.6$. Thus

$$-19.6=0(t)-\frac12(9.8)t^2.$$

Hence

$$t^2=4,$$

and the physical root is

$$t=2\ \mathrm s.$$

The horizontal distance is

$$x=12(2)=24\ \mathrm m.$$

At impact,

$$v_x=12, \qquad v_y=0-9.8(2)=-19.6.$$

Therefore the impact speed is

$$|\mathbf v|=\sqrt{12^2+19.6^2} \approx23.0\ \mathrm{m\,s^{-1}}.$$

The answers are

$$\boxed{2\ \mathrm s, \qquad24\ \mathrm m, \qquad23.0\ \mathrm{m\,s^{-1}}}.$$

Notice that the time to fall was determined entirely by the vertical motion. A greater horizontal launch speed would increase the range but not the time to hit the ground in this model.

Self-check 4

A ball is projected horizontally at $8\ \mathrm{m\,s^{-1}}$ from a window $11.025$ m above level ground. Find the time before impact and the horizontal distance travelled. Use $g=9.8\ \mathrm{m\,s^{-2}}$.

Answer

Vertically,

$$-11.025=-\frac12(9.8)t^2,$$

so $t^2=2.25$ and

$$\boxed{t=1.5\ \mathrm s}.$$

Horizontally,

$$\boxed{x=8(1.5)=12\ \mathrm m}.$$

Hitting a target

For a target with known coordinates, the horizontal equation often gives the arrival time immediately. Substitute that time into the vertical equation and compare the calculated height with the target height.

Worked example 6: clearing a wall

A ball is projected from ground level at $18\ \mathrm{m\,s^{-1}}$ at $40^\circ$ above the horizontal. A vertical wall of height $6$ m is $20$ m from the launch point. Determine whether the ball clears the wall.

At the wall, $x=20$. From horizontal motion,

$$20=18\cos40^\circ\,t,$$

so

$$t=\frac{20}{18\cos40^\circ} \approx1.4505\ \mathrm s.$$

Use this unrounded time vertically:

$$\begin{aligned} y &=18\sin40^\circ\,t-4.9t^2\\ &\approx6.39\ \mathrm m. \end{aligned}$$

Since $6.39>6$, the ball clears the wall by approximately

$$\boxed{0.39\ \mathrm m}.$$

The conclusion is about the particle model. If the ball has appreciable radius, its centre would need to clear a slightly greater height.

Worked example 7: finding an unknown launch speed

A particle is projected at angle $45^\circ$ above the horizontal and passes through the point $(30,10)$, where distances are in metres. Find the launch speed $U$.

Use the trajectory equation:

$$10=30\tan45^\circ -\frac{9.8(30)^2}{2U^2\cos^245^\circ}.$$

Since $\tan45^\circ=1$ and $\cos^245^\circ=1/2$,

$$10=30-\frac{8820}{U^2}.$$

Therefore

$$\frac{8820}{U^2}=20, \qquad U^2=441.$$

Speed is positive, so

$$\boxed{U=21\ \mathrm{m\,s^{-1}}}.$$

Self-check 5

A projectile is launched from the origin with initial velocity $(10\mathbf i+14\mathbf j)\ \mathrm{m\,s^{-1}}$. Does it pass above or below the point $(15,8)$? By how much?

Answer

At $x=15$,

$$15=10t \quad\Longrightarrow\quad t=1.5\ \mathrm s.$$

Its height then is

$$y=14(1.5)-4.9(1.5)^2=9.975\ \mathrm m.$$

Therefore it passes

$$9.975-8=\boxed{1.975\ \mathrm m}$$

above the point.

Common misconceptions

Gravity affects horizontal motion

In the standard model, gravity acts vertically. Therefore $a_x=0$, not $-g$, and horizontal velocity stays constant.

Acceleration is zero at maximum height

At maximum height, $v_y=0$ for one instant, but $a_y=-g$ throughout the flight. The projectile immediately begins to descend.

The projectile has zero speed at maximum height

Only the vertical component is zero. Unless the projection was vertical, the speed there is $|U\cos\theta|$.

A negative quadratic root is another physical event

A negative time usually describes the mathematical extension of the trajectory before launch. State and keep only roots belonging to the modelled flight.

The range formula always applies

The formula $R=U^2\sin2\theta/g$ requires launch and landing at the same height. For unequal levels, solve the vertical displacement equation first.

Sine is always vertical

The component adjacent to the marked angle uses cosine. If the angle is measured from the horizontal, $U\cos\theta$ is horizontal. If it is measured from the vertical, the assignments reverse.

Mixed self-check

A particle is projected from a point $5$ m above level ground with speed $20\ \mathrm{m\,s^{-1}}$ at $30^\circ$ above the horizontal. Use $g=9.8\ \mathrm{m\,s^{-2}}$.

1. Find the time when it first reaches a height of $8$ m above the ground.
2. Find its maximum height above the ground.
3. Find the time at which it hits the ground.
4. Find its horizontal range.

Answers

Taking ground as $y=0$,

$$y=5+10t-4.9t^2, \qquad x=20\cos30^\circ\,t=10\sqrt3\,t.$$

For $y=8$,

$$8=5+10t-4.9t^2,$$

so

$$4.9t^2-10t+3=0.$$

Thus

$$t=\frac{10\pm\sqrt{100-58.8}}{9.8} =\frac{10\pm\sqrt{41.2}}{9.8}.$$

The first time is

$$\boxed{t\approx0.365\ \mathrm s}.$$

The second root, approximately $1.68$ s, is when it passes the same height while descending.

The rise above the launch point is

$$\frac{10^2}{2(9.8)}=\frac{250}{49}\ \mathrm m.$$

Therefore the maximum height above ground is

$$\boxed{5+\frac{250}{49}\approx10.1\ \mathrm m}.$$

At ground level,

$$0=5+10t-4.9t^2.$$

The positive root is

$$t=\frac{10+\sqrt{198}}{9.8} \approx2.46\ \mathrm s.$$

Hence the range is

$$x=10\sqrt3\left(\frac{10+\sqrt{198}}{9.8}\right) \approx\boxed{42.5\ \mathrm m}.$$

Summary

For projection speed $U$ at angle $\theta$ above the horizontal, from the origin,

$$\begin{aligned} x&=U\cos\theta\,t, &v_x&=U\cos\theta,\\ y&=U\sin\theta\,t-\frac12gt^2, &v_y&=U\sin\theta-gt. \end{aligned}$$

At maximum height, set $v_y=0$. At a specified vertical level, set $y$ equal to that displacement. At a specified horizontal position, use $x=U\cos\theta\,t$ to find the shared time.

For return to the launch level only,

$$T=\frac{2U\sin\theta}{g}, \qquad R=\frac{U^2\sin2\theta}{g}.$$

The trajectory equation is

$$y=x\tan\theta-\frac{gx^2}{2U^2\cos^2\theta}.$$

Next, connect component accelerations to their causes in dynamics in a plane, or strengthen the vector foundation in two dimensional motion.