Projectile Calculations | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Projectile Calculations

Every projectile calculation follows the same four steps. Time is shared by both directions: use whichever direction contains enough known information to find it.

The method

Split the launch velocity: $u_x = u\cos\theta$ , $u_y = u\sin\theta$ .
Vertical: use SUVAT with downward acceleration $g$ to find a time, height or vertical velocity.
Horizontal: use $x = u_x t$ . It can also provide time when horizontal distance is given.
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 $u_y = 0$ .
Height: $h = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(9.81)(0.81)^2 = 3.2 \text{ m}$ .
Distance: $d = u_x t = 5.4 \times 0.81 = 4.4 \text{ m}$ .

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: $u_x = 22\cos 40^\circ = 16.9 \text{ m s}^{-1}$ , $u_y = 22\sin 40^\circ = 14 \text{ m s}^{-1}$ upward.
Time to reach the wall (horizontal): $t = 36/16.9 = 2.1 \text{ s}$ .
Vertical displacement in that time (up positive):
$ut=14(2.1)=29.4\ \text{m}$
$\tfrac{1}{2}gt^2=21.6\ \text{m}$
$s=29.4-21.6=7.8\ \text{m}$
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 turnAdapted from 9702/23/M/J/25 Q1(c) · 3 marks

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))

Show worked answer

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 vertical landing component is unchanged, but the new throw also has a horizontal component that remains constant. Combining the two perpendicular components gives agreater landing speed.

Answer

The time is unchanged because the initial vertical component is unchanged. The landing speed is greater because the new horizontal component remains throughout the flight while the vertical landing component is unchanged.

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.

Projectile Calculations

Every projectile calculation follows the same four steps. Time is shared by both directions: use whichever direction contains enough known information to find it.

The method

Split the launch velocity: $u_x = u\cos\theta$ , $u_y = u\sin\theta$ .
Vertical: use SUVAT with downward acceleration $g$ to find a time, height or vertical velocity.
Horizontal: use $x = u_x t$ . It can also provide time when horizontal distance is given.
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 $u_y = 0$ .
Height: $h = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(9.81)(0.81)^2 = 3.2 \text{ m}$ .
Distance: $d = u_x t = 5.4 \times 0.81 = 4.4 \text{ m}$ .

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: $u_x = 22\cos 40^\circ = 16.9 \text{ m s}^{-1}$ , $u_y = 22\sin 40^\circ = 14 \text{ m s}^{-1}$ upward.
Time to reach the wall (horizontal): $t = 36/16.9 = 2.1 \text{ s}$ .
Vertical displacement in that time (up positive):
$ut=14(2.1)=29.4\ \text{m}$
$\tfrac{1}{2}gt^2=21.6\ \text{m}$
$s=29.4-21.6=7.8\ \text{m}$
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 turnAdapted from 9702/23/M/J/25 Q1(c) · 3 marks

Show worked answer

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 vertical landing component is unchanged, but the new throw also has a horizontal component that remains constant. Combining the two perpendicular components gives agreater landing speed.

Answer

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.