Physics 9702 · Paper 1

Projectile Calculations

Every projectile calculation follows the same four steps, and the vertical motion almost always provides the time.

The method

Split the launch velocity: $u_H = v\cos\theta$ , $u_V = v\sin\theta$ .
Vertical: use SUVAT with $a = g$ to find the time (or a height).
Horizontal: use $x = u_H t$ with that same time.
If the question asks for a height above the ground, add the release height at the end.

Worked example

Warm-up: a horizontal launch

An aircraft releases a parcel horizontally at 5.4 m s⁻¹. The parcel reaches the ground after 0.81 s. Find the release height and the horizontal distance travelled.

A horizontal release means $u_V = 0$ .
Height: $h = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(9.81)(0.81)^2 = 3.2 \text{ m}$ .
Distance: $d = u_H t = 5.4 \times 0.81 = 4.4 \text{ m}$ .

Answer

h = 3.2 \text{ m}, \quad d = 4.4 \text{ m}

(9702/21/M/J/20 Q2(b))

Worked example

Exam-style: an angled throw over a wall

A ball is thrown at 22 m s⁻¹ at 40° above the horizontal, from 1.2 m above the ground. A wall stands 36 m away. Find the height of the ball when it reaches the wall.

Split the launch once: 22 m s⁻¹ at 40° gives 16.9 m s⁻¹ across and 14 m s⁻¹ up. The 36 m across fixes the time; the vertical motion gives the 9.0 m height at the wall.

Split: $u_H = 22\cos 40^\circ = 16.9 \text{ m s}^{-1}$ , $u_V = 22\sin 40^\circ = 14 \text{ m s}^{-1}$ upward.
Time to reach the wall (horizontal): $t = 36/16.9 = 2.1 \text{ s}$ .
Vertical displacement in that time (up positive):
$s = 14(2.1) - \tfrac{1}{2}(9.81)(2.1)^2 = 29.4 - 21.6 = 7.8 \text{ m}$
Height above the ground: $h = 7.8 + 1.2 = 9.0 \text{ m}$ .

Answer

h = 9.0 \text{ m}

(9702/21/M/J/24 Q2(b)(iii))

Common mistake

SUVAT gives the displacement above the release point, not above the ground. When the throw starts above the ground, add the release height as the final step. Forgetting it loses the last mark.

The explain variants

Newer papers also ask projectile questions with no calculation: you compare two situations and explain, using the independence idea and energy.

Your turn— tap to reveal the worked answer (9702/23/M/J/25 Q1(c))

Object B is thrown from a building with an upward vertical velocity of 3.0 m s⁻¹ and reaches the ground in time $t$ . It is then thrown from the same height at 6.0 m s⁻¹ at 60° to the vertical. State and explain how the new time to reach the ground, and the new landing speed, compare. (9702/23/M/J/25 Q1(c))

Time: the vertical component is $6.0\cos 60^\circ = 3.0 \text{ m s}^{-1}$ upward, the same as before. The vertical motion is identical, so the time is the same. The extra horizontal motion does not affect the fall.
Landing speed: the object now starts with more kinetic energy (speed 6.0 instead of 3.0 m s⁻¹) and loses the same gravitational potential energy, so it lands faster.

The variant map for projectiles: horizontal launch (the easiest), angled launch to a wall or target, time-comparison explain questions like the one above, and component questions (lesson 2.8). All of them start with the same split.

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Projectile Calculations

Every projectile calculation follows the same four steps, and the vertical motion almost always provides the time.

The method

Split the launch velocity: $u_H = v\cos\theta$ , $u_V = v\sin\theta$ .
Vertical: use SUVAT with $a = g$ to find the time (or a height).
Horizontal: use $x = u_H t$ with that same time.
If the question asks for a height above the ground, add the release height at the end.

Worked example

Warm-up: a horizontal launch

An aircraft releases a parcel horizontally at 5.4 m s⁻¹. The parcel reaches the ground after 0.81 s. Find the release height and the horizontal distance travelled.

A horizontal release means $u_V = 0$ .
Height: $h = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(9.81)(0.81)^2 = 3.2 \text{ m}$ .
Distance: $d = u_H t = 5.4 \times 0.81 = 4.4 \text{ m}$ .

Answer

h = 3.2 \text{ m}, \quad d = 4.4 \text{ m}

(9702/21/M/J/20 Q2(b))

Worked example

Exam-style: an angled throw over a wall

A ball is thrown at 22 m s⁻¹ at 40° above the horizontal, from 1.2 m above the ground. A wall stands 36 m away. Find the height of the ball when it reaches the wall.

Split the launch once: 22 m s⁻¹ at 40° gives 16.9 m s⁻¹ across and 14 m s⁻¹ up. The 36 m across fixes the time; the vertical motion gives the 9.0 m height at the wall.

Split: $u_H = 22\cos 40^\circ = 16.9 \text{ m s}^{-1}$ , $u_V = 22\sin 40^\circ = 14 \text{ m s}^{-1}$ upward.
Time to reach the wall (horizontal): $t = 36/16.9 = 2.1 \text{ s}$ .
Vertical displacement in that time (up positive):
$s = 14(2.1) - \tfrac{1}{2}(9.81)(2.1)^2 = 29.4 - 21.6 = 7.8 \text{ m}$
Height above the ground: $h = 7.8 + 1.2 = 9.0 \text{ m}$ .

Answer

h = 9.0 \text{ m}

(9702/21/M/J/24 Q2(b)(iii))

Common mistake

SUVAT gives the displacement above the release point, not above the ground. When the throw starts above the ground, add the release height as the final step. Forgetting it loses the last mark.

The explain variants

Newer papers also ask projectile questions with no calculation: you compare two situations and explain, using the independence idea and energy.

Your turn— tap to reveal the worked answer (9702/23/M/J/25 Q1(c))

Time: the vertical component is $6.0\cos 60^\circ = 3.0 \text{ m s}^{-1}$ upward, the same as before. The vertical motion is identical, so the time is the same. The extra horizontal motion does not affect the fall.
Landing speed: the object now starts with more kinetic energy (speed 6.0 instead of 3.0 m s⁻¹) and loses the same gravitational potential energy, so it lands faster.

The variant map for projectiles: horizontal launch (the easiest), angled launch to a wall or target, time-comparison explain questions like the one above, and component questions (lesson 2.8). All of them start with the same split.