Projectiles

What you'll learn

How to split a launch velocity into horizontal and vertical components.
How to use SUVAT separately in the two directions.
How to find greatest height, time of flight, range, and speed.
How to handle raised launch points, trajectory equations, and collisions.

The projectile model

A projectile is an object that has been launched and then moves only under gravity.

Definition

Projectile model

In the standard A-Level model, the object is treated as a particle, air resistance is ignored, and the acceleration is constant: vertically downwards with magnitude $g$ , where $g = 9.8$ m s⁻².

This means:

horizontal acceleration is zero
vertical acceleration is $- g$ if you take upwards as positive
horizontal and vertical motion happen at the same time, but you calculate them separately

Key Idea

Separate the motion

Projectile questions become much easier when you treat the horizontal direction and vertical direction as two linked SUVAT problems with the same time $t$ .

Splitting the initial velocity

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

The initial velocity can be resolved into horizontal and vertical components using right-angled trigonometry.

horizontal component: $Ucos⁡θU\cos\theta$
vertical component: $Usin⁡θU\sin\theta$

If the velocity is given as $ai+bja\mathbf{i}+b\mathbf{j}$ , then:

$i\mathbf{i}$ is the horizontal unit vector
$j\mathbf{j}$ is the vertical unit vector
initial horizontal velocity is $a$
initial vertical velocity is $b$

Core equations

Taking the launch point as the origin and upwards as positive:

x = u_x t

u_y t - \frac{1}{2}gt^2

v_x = u_x

v_y = u_y - gt

where $u_x$ and $u_y$ are the initial horizontal and vertical components.

Example

Velocity after a given time

A particle is projected with speed 18 m s⁻¹ at an angle of $50∘50^\circ$ above the horizontal. Find its velocity after 2 seconds.

Resolve the initial velocity into components:
$ux=18cos⁡50∘=11.57u_x = 18\cos 50^\circ = 11.57$ $uy=18sin⁡50∘=13.79u_y = 18\sin 50^\circ = 13.79$
The horizontal velocity stays constant:
$v_x = 11.57$
Use vertical motion with acceleration $- 9.8$ :
$v_y = 13.79 - 9.8(2) = -5.81$
Write the velocity as a vector:
$v=11.6i−5.81j\mathbf{v} = 11.6\mathbf{i} - 5.81\mathbf{j}$

Common Mistake

Putting gravity in the horizontal direction

Do not use $g$ in the horizontal equation. Gravity acts vertically, so the horizontal velocity is constant unless air resistance is included.

Greatest height

At the greatest height, the vertical velocity is zero. The particle is still moving horizontally, but momentarily it is not moving up or down.

At the top of the flight the vertical velocity is zero but the horizontal velocity remains non-zero.

Use vertical SUVAT only:

v_y^2 = u_y^2 + 2as

At the top, $v_y=0$ , $a = - g$ , and $s = H$ , where $H$ is the height gained above the launch point.

Example

Finding the greatest height

A particle is projected from level ground with speed 22 m s⁻¹ at an angle of $50∘50^\circ$ to the horizontal. Find its greatest height above the ground.

Find the initial vertical velocity:
$uy=22sin⁡50∘=16.85u_y = 22\sin 50^\circ = 16.85$
At the highest point, the vertical velocity is zero:
$0 = 16.85^2 - 2(9.8)H$
Solve for $H$ :
$\frac{16.85^2}{19.6} = 14.5$
The greatest height is 14.5 m.

Tip

Top of the flight

At maximum height, only the vertical velocity is zero. The horizontal velocity is not zero.

Time of flight and horizontal range

The range is the horizontal distance travelled before the projectile lands.

For level-ground projection, the range is the horizontal distance from launch to landing at the same height.

Definition

Range

The range of a projectile is the horizontal distance from its launch point to its landing point.

If the particle lands at the same height it was launched from, set $y = 0$ and solve for the non-zero value of $t$ .

Then use:

x = u_x t

Example

Finding the range on level ground

A ball is projected from the ground with speed 22 m s⁻¹ at an angle of $50∘50^\circ$ . Find how far away it lands.

Use the same components:
$ux=22cos⁡50∘=14.14u_x = 22\cos 50^\circ = 14.14$ $uy=22sin⁡50∘=16.85u_y = 22\sin 50^\circ = 16.85$
Since it lands back on the ground, vertical displacement is zero:
$0 = 16.85t - 4.9t^2$
Factorise to find the non-zero time:
$t (16.85 - 4.9 t) = 0$ $\frac{16.85}{4.9} = 3.44$
Use horizontal motion to find the range:
$x = 14.14 (3.44) = 48.6$
The horizontal distance is 48.6 m.

Common Mistake

Range formula limitation

The shortcut $R=U2sin⁡2θgR=\frac{U^2\sin 2\theta}{g}$ only works when the projectile lands at the same vertical height as it was launched.

Starting above the ground

If the projectile starts at height $h$ above the ground, use:

A raised launch point requires height above ground to be measured from the ground, not from the launch point alone.

ground=h+uyt−12gt2\text{height above ground} = h + u_y t - \frac{1}{2}gt^2

To find when it hits the ground, set the height equal to zero.

Example

Projection from a raised point

A particle is projected from a point 12 m above horizontal ground with velocity $5i+8j5\mathbf{i}+8\mathbf{j}$ m s⁻¹. Find its greatest height above the ground and the horizontal distance travelled before it lands.

Read off the initial components:
$ux=5,uy=8u_x = 5,\qquad u_y = 8$
Find the extra height gained above the launch point:
$\frac{8^2}{2(9.8)} = 3.27$
Add the starting height:
$ground=12+3.27=15.3\text{greatest height above ground} = 12 + 3.27 = 15.3$
Set the height above ground equal to zero:
$0 = 12 + 8t - 4.9t^2$
Rearrange and solve the quadratic, taking the positive root:
$4.9t^2 - 8t - 12 = 0$ $t = 2.58$
Use horizontal motion:
$x = 5 (2.58) = 12.9$
The particle travels 12.9 m horizontally before landing.

Finding unknown speed or angle

Sometimes you are given the time of flight, greatest height, or a point on the path, and you need to find $U$ or $θ\theta$ .

A useful method is to set up equations using the components:

ux=Ucos⁡θu_x = U\cos\theta

uy=Usin⁡θu_y = U\sin\theta

Example

Finding an unknown angle from the flight time

A particle is projected from the ground with speed 30 m s⁻¹ at angle $θ\theta$ above the horizontal. It lands back on the ground after 4 seconds. Find $θ\theta$ and the range.

Use vertical displacement equal to zero:
$(30\sin\theta)(4) - 4.9(4^2)$
Simplify:
$120sin⁡θ=78.4120\sin\theta = 78.4$
Find $θ\theta$ :
$sin⁡θ=0.6533\sin\theta = 0.6533$ $θ=40.8∘\theta = 40.8^\circ$
Use horizontal motion:
$(30\cos 40.8^\circ)(4)$
Calculate the range:
$x = 90.9$
The range is 90.9 m.

The trajectory equation

The trajectory is the path followed by the projectile. You can remove time from the equations to connect horizontal distance $x$ and vertical displacement $y$ directly.

The trajectory equation gives the vertical displacement y directly in terms of horizontal distance x.

From horizontal motion:

\frac{x}{U\cos\theta}

Substitute this into the vertical equation:

x\tan\theta - \frac{gx^2}{2U^2\cos^2\theta}

This is especially useful when a projectile passes through a known point.

Example

Using a point on the path to find the launch speed

A ball is projected at $45∘45^\circ$ from a point 10 m above the ground. It lands on the ground 60 m horizontally from the point below launch. Find the launch speed.

The landing point is 60 m horizontally from the foot of the launch point and 10 m below the launch level.

Work relative to the launch point. At landing:
$60,\qquad y = -10$
Use the trajectory equation with $θ=45∘\theta=45^\circ$ :
$60\tan45^\circ - \frac{9.8(60^2)}{2U^2\cos^2 45^\circ}$
Use $tan⁡45∘=1\tan45^\circ=1$ and $cos⁡245∘=12\cos^2 45^\circ=\frac{1}{2}$ :
$\frac{9.8(3600)}{U^2}$
Solve for $U^2$ :
$35280U2=70\frac{35280}{U^2} = 70$ $U^2 = 504$
Find $U$ :
$\sqrt{504} = 22.4$
The launch speed is 22.4 m s⁻¹.

Collisions between two projectiles

For two projectiles to collide, they must have the same position at the same time.

Choose a coordinate system carefully. A common choice is:

origin at the left-hand launch point
positive horizontal direction to the right
positive vertical direction upwards

Example

Two balls colliding

Two points A and B are 40 m apart on horizontal ground. Ball P is projected from A with speed 20 m s⁻¹ at $30∘30^\circ$ above the horizontal towards B. Ball Q is projected from B towards A and collides with P after 2 seconds. Find the velocity of P just before collision, and find the initial speed and angle of Q.

For a collision, both projectiles must occupy the same point at the same time in the chosen coordinate system.

Resolve P’s initial velocity:
$uPx=20cos⁡30∘=17.32u_{Px}=20\cos30^\circ=17.32$ $uPy=20sin⁡30∘=10u_{Py}=20\sin30^\circ=10$
Find P’s velocity at $t = 2$ :
$v_{Px}=17.32$ $v_{Py}=10-9.8(2)=-9.6$
So P’s velocity just before collision is:
$vP=17.3i−9.6j\mathbf{v}_P = 17.3\mathbf{i}-9.6\mathbf{j}$
Find P’s position at $t = 2$ :
$x_P=17.32(2)=34.64$ $y_P=10(2)-4.9(2^2)=0.4$
Ball Q starts at $x = 40$ , so its horizontal displacement is:
$34.64 - 40 = - 5.36$
Therefore Q’s horizontal component towards A has magnitude:
$Ucos⁡θ=5.362=2.68U\cos\theta=\frac{5.36}{2}=2.68$
Use Q’s vertical displacement:
$(U\sin\theta)(2)-4.9(2^2)$ $Usin⁡θ=10U\sin\theta=10$
Find $θ\theta$ :
$tan⁡θ=102.68\tan\theta=\frac{10}{2.68}$ $θ=75.0∘\theta=75.0^\circ$
Find $U$ :
$U=2.682+102=10.4U=\sqrt{2.68^2+10^2}=10.4$
Q is projected at 10.4 m s⁻¹ at an angle of $75.0∘75.0^\circ$ above the horizontal, towards A.

Exam technique

In the exam

Draw a quick diagram and choose positive directions before writing equations.
Resolve the initial velocity first; most errors happen before SUVAT even begins.
Use vertical motion for height and time, then horizontal motion for distance.
If the launch and landing heights differ, do not use the level-ground range formula.
For collisions, make the two position vectors equal at the same time.

Self review

Check yourself

Why is the horizontal acceleration zero in the standard projectile model?
At the greatest height, which component of velocity is zero?
If a projectile starts 15 m above the ground, what vertical displacement should you use when it lands?

Projectiles

What you'll learn

How to split a launch velocity into horizontal and vertical components.
How to use SUVAT separately in the two directions.
How to find greatest height, time of flight, range, and speed.
How to handle raised launch points, trajectory equations, and collisions.

The projectile model

A projectile is an object that has been launched and then moves only under gravity.

Definition

Projectile model

In the standard A-Level model, the object is treated as a particle, air resistance is ignored, and the acceleration is constant: vertically downwards with magnitude $g$ , where $g = 9.8$ m s⁻².

This means:

horizontal acceleration is zero
vertical acceleration is $- g$ if you take upwards as positive
horizontal and vertical motion happen at the same time, but you calculate them separately

Key Idea

Separate the motion

Projectile questions become much easier when you treat the horizontal direction and vertical direction as two linked SUVAT problems with the same time $t$ .

Splitting the initial velocity

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

The initial velocity can be resolved into horizontal and vertical components using right-angled trigonometry.

horizontal component: $Ucos⁡θU\cos\theta$
vertical component: $Usin⁡θU\sin\theta$

If the velocity is given as $ai+bja\mathbf{i}+b\mathbf{j}$ , then:

$i\mathbf{i}$ is the horizontal unit vector
$j\mathbf{j}$ is the vertical unit vector
initial horizontal velocity is $a$
initial vertical velocity is $b$

Core equations

Taking the launch point as the origin and upwards as positive:

x = u_x t

u_y t - \frac{1}{2}gt^2

v_x = u_x

v_y = u_y - gt

where $u_x$ and $u_y$ are the initial horizontal and vertical components.

Example

Velocity after a given time

A particle is projected with speed 18 m s⁻¹ at an angle of $50∘50^\circ$ above the horizontal. Find its velocity after 2 seconds.

Resolve the initial velocity into components:
$ux=18cos⁡50∘=11.57u_x = 18\cos 50^\circ = 11.57$ $uy=18sin⁡50∘=13.79u_y = 18\sin 50^\circ = 13.79$
The horizontal velocity stays constant:
$v_x = 11.57$
Use vertical motion with acceleration $- 9.8$ :
$v_y = 13.79 - 9.8(2) = -5.81$
Write the velocity as a vector:
$v=11.6i−5.81j\mathbf{v} = 11.6\mathbf{i} - 5.81\mathbf{j}$

Common Mistake

Putting gravity in the horizontal direction

Do not use $g$ in the horizontal equation. Gravity acts vertically, so the horizontal velocity is constant unless air resistance is included.

Greatest height

At the greatest height, the vertical velocity is zero. The particle is still moving horizontally, but momentarily it is not moving up or down.

At the top of the flight the vertical velocity is zero but the horizontal velocity remains non-zero.

Use vertical SUVAT only:

v_y^2 = u_y^2 + 2as

At the top, $v_y=0$ , $a = - g$ , and $s = H$ , where $H$ is the height gained above the launch point.

Example

Finding the greatest height

A particle is projected from level ground with speed 22 m s⁻¹ at an angle of $50∘50^\circ$ to the horizontal. Find its greatest height above the ground.

Find the initial vertical velocity:
$uy=22sin⁡50∘=16.85u_y = 22\sin 50^\circ = 16.85$
At the highest point, the vertical velocity is zero:
$0 = 16.85^2 - 2(9.8)H$
Solve for $H$ :
$\frac{16.85^2}{19.6} = 14.5$
The greatest height is 14.5 m.

Tip

Top of the flight

At maximum height, only the vertical velocity is zero. The horizontal velocity is not zero.

Time of flight and horizontal range

The range is the horizontal distance travelled before the projectile lands.

For level-ground projection, the range is the horizontal distance from launch to landing at the same height.

Definition

Range

The range of a projectile is the horizontal distance from its launch point to its landing point.

If the particle lands at the same height it was launched from, set $y = 0$ and solve for the non-zero value of $t$ .

Then use:

x = u_x t

Example

Finding the range on level ground

A ball is projected from the ground with speed 22 m s⁻¹ at an angle of $50∘50^\circ$ . Find how far away it lands.

Use the same components:
$ux=22cos⁡50∘=14.14u_x = 22\cos 50^\circ = 14.14$ $uy=22sin⁡50∘=16.85u_y = 22\sin 50^\circ = 16.85$
Since it lands back on the ground, vertical displacement is zero:
$0 = 16.85t - 4.9t^2$
Factorise to find the non-zero time:
$t (16.85 - 4.9 t) = 0$ $\frac{16.85}{4.9} = 3.44$
Use horizontal motion to find the range:
$x = 14.14 (3.44) = 48.6$
The horizontal distance is 48.6 m.

Common Mistake

Range formula limitation

The shortcut $R=U2sin⁡2θgR=\frac{U^2\sin 2\theta}{g}$ only works when the projectile lands at the same vertical height as it was launched.

Starting above the ground

If the projectile starts at height $h$ above the ground, use:

A raised launch point requires height above ground to be measured from the ground, not from the launch point alone.

ground=h+uyt−12gt2\text{height above ground} = h + u_y t - \frac{1}{2}gt^2

To find when it hits the ground, set the height equal to zero.

Example

Projection from a raised point

Read off the initial components:
$ux=5,uy=8u_x = 5,\qquad u_y = 8$
Find the extra height gained above the launch point:
$\frac{8^2}{2(9.8)} = 3.27$
Add the starting height:
$ground=12+3.27=15.3\text{greatest height above ground} = 12 + 3.27 = 15.3$
Set the height above ground equal to zero:
$0 = 12 + 8t - 4.9t^2$
Rearrange and solve the quadratic, taking the positive root:
$4.9t^2 - 8t - 12 = 0$ $t = 2.58$
Use horizontal motion:
$x = 5 (2.58) = 12.9$
The particle travels 12.9 m horizontally before landing.

Finding unknown speed or angle

Sometimes you are given the time of flight, greatest height, or a point on the path, and you need to find $U$ or $θ\theta$ .

A useful method is to set up equations using the components:

ux=Ucos⁡θu_x = U\cos\theta

uy=Usin⁡θu_y = U\sin\theta

Example

Finding an unknown angle from the flight time

A particle is projected from the ground with speed 30 m s⁻¹ at angle $θ\theta$ above the horizontal. It lands back on the ground after 4 seconds. Find $θ\theta$ and the range.

Use vertical displacement equal to zero:
$(30\sin\theta)(4) - 4.9(4^2)$
Simplify:
$120sin⁡θ=78.4120\sin\theta = 78.4$
Find $θ\theta$ :
$sin⁡θ=0.6533\sin\theta = 0.6533$ $θ=40.8∘\theta = 40.8^\circ$
Use horizontal motion:
$(30\cos 40.8^\circ)(4)$
Calculate the range:
$x = 90.9$
The range is 90.9 m.

The trajectory equation

The trajectory is the path followed by the projectile. You can remove time from the equations to connect horizontal distance $x$ and vertical displacement $y$ directly.

The trajectory equation gives the vertical displacement y directly in terms of horizontal distance x.

From horizontal motion:

\frac{x}{U\cos\theta}

Substitute this into the vertical equation:

x\tan\theta - \frac{gx^2}{2U^2\cos^2\theta}

This is especially useful when a projectile passes through a known point.

Example

Using a point on the path to find the launch speed

A ball is projected at $45∘45^\circ$ from a point 10 m above the ground. It lands on the ground 60 m horizontally from the point below launch. Find the launch speed.

The landing point is 60 m horizontally from the foot of the launch point and 10 m below the launch level.

Work relative to the launch point. At landing:
$60,\qquad y = -10$
Use the trajectory equation with $θ=45∘\theta=45^\circ$ :
$60\tan45^\circ - \frac{9.8(60^2)}{2U^2\cos^2 45^\circ}$
Use $tan⁡45∘=1\tan45^\circ=1$ and $cos⁡245∘=12\cos^2 45^\circ=\frac{1}{2}$ :
$\frac{9.8(3600)}{U^2}$
Solve for $U^2$ :
$35280U2=70\frac{35280}{U^2} = 70$ $U^2 = 504$
Find $U$ :
$\sqrt{504} = 22.4$
The launch speed is 22.4 m s⁻¹.

Collisions between two projectiles

For two projectiles to collide, they must have the same position at the same time.

Choose a coordinate system carefully. A common choice is:

origin at the left-hand launch point
positive horizontal direction to the right
positive vertical direction upwards

Example

Two balls colliding

For a collision, both projectiles must occupy the same point at the same time in the chosen coordinate system.

Resolve P’s initial velocity:
$uPx=20cos⁡30∘=17.32u_{Px}=20\cos30^\circ=17.32$ $uPy=20sin⁡30∘=10u_{Py}=20\sin30^\circ=10$
Find P’s velocity at $t = 2$ :
$v_{Px}=17.32$ $v_{Py}=10-9.8(2)=-9.6$
So P’s velocity just before collision is:
$vP=17.3i−9.6j\mathbf{v}_P = 17.3\mathbf{i}-9.6\mathbf{j}$
Find P’s position at $t = 2$ :
$x_P=17.32(2)=34.64$ $y_P=10(2)-4.9(2^2)=0.4$
Ball Q starts at $x = 40$ , so its horizontal displacement is:
$34.64 - 40 = - 5.36$
Therefore Q’s horizontal component towards A has magnitude:
$Ucos⁡θ=5.362=2.68U\cos\theta=\frac{5.36}{2}=2.68$
Use Q’s vertical displacement:
$(U\sin\theta)(2)-4.9(2^2)$ $Usin⁡θ=10U\sin\theta=10$
Find $θ\theta$ :
$tan⁡θ=102.68\tan\theta=\frac{10}{2.68}$ $θ=75.0∘\theta=75.0^\circ$
Find $U$ :
$U=2.682+102=10.4U=\sqrt{2.68^2+10^2}=10.4$
Q is projected at 10.4 m s⁻¹ at an angle of $75.0∘75.0^\circ$ above the horizontal, towards A.

Exam technique

In the exam

Draw a quick diagram and choose positive directions before writing equations.
Resolve the initial velocity first; most errors happen before SUVAT even begins.
Use vertical motion for height and time, then horizontal motion for distance.
If the launch and landing heights differ, do not use the level-ground range formula.
For collisions, make the two position vectors equal at the same time.

Self review

Check yourself

Why is the horizontal acceleration zero in the standard projectile model?
At the greatest height, which component of velocity is zero?
If a projectile starts 15 m above the ground, what vertical displacement should you use when it lands?

Projectiles

What you'll learn

The projectile model

Splitting the initial velocity

Core equations

Greatest height

Time of flight and horizontal range

Starting above the ground

Finding unknown speed or angle

The trajectory equation

Collisions between two projectiles

Recap questions

Projectiles

What you'll learn

The projectile model

Splitting the initial velocity

Core equations

Greatest height

Time of flight and horizontal range

Starting above the ground

Finding unknown speed or angle

The trajectory equation

Collisions between two projectiles

Recap questions

Projectiles

What you'll learn

The projectile model

Splitting the initial velocity

Core equations

Greatest height

Time of flight and horizontal range

Starting above the ground

Finding unknown speed or angle

The trajectory equation

Collisions between two projectiles

Recap questions

Projectiles Revision Guide

Projectiles

What you'll learn

The projectile model

Splitting the initial velocity

Core equations

Greatest height

Time of flight and horizontal range

Starting above the ground

Finding unknown speed or angle

The trajectory equation

Collisions between two projectiles

Recap questions

Projectiles Revision Guide