Vertical projectile motion — Physical Sciences

4 exam-style question sets on Vertical projectile motion (Paper 1), each with a hint and a fully worked answer. The app holds 20 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Two objects launched at different timesGraphs of motionThrow from a roof with velocity table & bounceObject released from a moving carrier

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Question 1

Two objects launched at different timesApplication16 marksPaper 1

A maintenance worker stands on the flat top of a water tower. He throws ball R vertically upwards at 11 m·s⁻¹ from the level of the top. Exactly 2 s after R is thrown, he throws ball S vertically downwards at 7 m·s⁻¹ from the same level. Both balls strike the ground at the SAME instant, at time t after R was thrown.

Ignore the effects of air friction.

The position-time sketch graphs of the two balls, with position measured from the ground, are shown. H is the height of the top of the tower above the ground and M is the maximum position reached by ball R.

aUsing EQUATIONS OF MOTION ONLY, calculate the time t. (5)
bCalculate the height H of the top of the tower above the ground. (3)
cCalculate the maximum height M reached by ball R above the ground. (4)
dOn the same set of axes, sketch the velocity-time graphs for balls R and S from the moment each is thrown until it strikes the ground. Label the graphs R and S. Clearly indicate on the graphs: - The initial velocity of each ball - The time at which ball S is thrown - The time t (4)

This question has a diagram, shown in the app.

Hint

Both balls undergo the same displacement −H; write each displacement with Δy = v·t − ½g t², using (t − 2) for ball S, and set them equal.

Worked answer

a) Take upwards as positive. Both balls end at displacement $- H$ measured from the top.
Ball R: $Δ y = 11 t - 4, 9 t^{2}$ ✓
Ball S: $Δ y = - 7 (t - 2) - 4, 9 (t - 2)^{2}$ ✓
Set equal: $11 t - 4, 9 t^{2} = - 7 (t - 2) - 4, 9 (t - 2)^{2}$ ✓
$11 t = 12, 6 t - 5, 6$ ✓ → $t = \frac{5 , 6}{1 , 6} = 3, 50$ s ✓

b) Ball S falls for $3, 5 - 2 = 1, 5$ s:
$Δ y = v_{i} t + \frac{1}{2} g t^{2}$ ✓ $= 7 (1, 5) + 4, 9 (1, 5)^{2}$ ✓ (downwards positive)
$H = 10, 5 + 11, 03 = 21, 53$ m ✓

c) Extra rise of R above the top ( $v_{f} = 0$ ):
$v_{f}^{2} = v_{i}^{2} - 2 g Δ y$ ✓ → $Δ y = \frac{1 1 ^{2}}{2 ( 9 , 8 )} = 6, 17$ m ✓
$M = H + 6, 17 = 21, 53 + 6, 17$ ✓ $= 27, 70$ m ✓

d) Two parallel straight lines with gradient $- 9, 8$ m·s⁻² ✓:

R starts at (0; 11), crosses zero at about 1,12 s, and ends at (3,5; −23,3) ✓
S starts at (2; −7) and ends at (3,5; −21,7) ✓
The values 11 and −7 shown on the velocity axis; 2 s and t = 3,5 s shown on the time axis ✓

Modelled on Nov 2024 Paper 1, Question 3 — same skills, new scenario.

Question 2

Graphs of motionComprehension2 marksPaper 1

A coin is flicked vertically downwards at 2 m·s⁻¹ from a high bridge and falls freely for t₁ seconds before striking the water. Air resistance is ignored. The sketch shows the acceleration-time graph for the fall.

What does the area between the graph line and the time axis, from 0 to t₁, represent?

This question has a diagram, shown in the app.

Hint

On an acceleration-time graph the area equals a × Δt — which kinematic quantity is that?

Worked answer

D. Area under an acceleration-time graph = a × Δt = Δv, the change in velocity. A is wrong because the coin already had an initial velocity of 2 m·s⁻¹, so the final velocity is Δv plus 2 m·s⁻¹. B is wrong: the average acceleration is read from the graph line itself (9,8 m·s⁻²), not its area. C is wrong: distance comes from the area under a *velocity*-time graph.

Modelled on Nov 2024 Paper 1, Question 1.2 — same skill, new scenario.

Question 3

Throw from a roof with velocity table & bounceApplication15 marksPaper 1

A ball is thrown vertically upwards at 10 m·s⁻¹ from the flat roof of a block of flats that is 15 m high. The ball misses the roof on the way down and strikes the ground 3,05 s after being thrown. It then bounces and reaches a maximum height of 9 m above the ground. Ignore the effects of air resistance.

The table below shows the magnitude of the ball's velocity at three instants before the bounce.

Time (s)	Magnitude of velocity (m·s⁻¹)
0	10
k	0
3,05	n

aDefine the term projectile. (2)
bUsing EQUATIONS OF MOTION ONLY, calculate the value of k. (3)
cCalculate the value of n. (3)
dIs the collision of the ball with the ground ELASTIC or INELASTIC? Explain the answer WITHOUT using calculations. (3)
eSketch a velocity versus time graph (take upwards as positive) for the motion of the ball from the moment it is thrown until it reaches its maximum height after the bounce. Show the following NUMERICAL VALUES on the graph: - The initial velocity - Time k - Velocity n (4)

This question has a diagram, shown in the app.

Hint

At the highest point the velocity is zero; use v = v_i + aΔt for both unknowns, keeping a consistent sign convention.

Worked answer

a) A projectile is an object upon which the only force acting is the gravitational force (it moves under the influence of gravity alone). ✓✓

b) Take upwards as positive. At the highest point $v_{f} = 0$ :
$v_{f} = v_{i} + a Δ t$ ✓
$0 = 10 + (- 9, 8) k$ ✓
$k = 1, 02$ s ✓

c) $v_{f} = v_{i} + a Δ t$ ✓
$v_{f} = 10 + (- 9, 8) (3, 05) = - 19, 89$ m·s⁻¹ ✓
$n = 19, 89$ m·s⁻¹ (magnitude) ✓

d) INELASTIC ✓ — after the bounce the ball rises to only 9 m, far lower than the height that corresponds to its impact speed, so it leaves the ground with a smaller speed than it struck the ground. ✓ The total kinetic energy therefore decreases during the collision, which makes it inelastic. ✓

e) Full-mark graph: a straight line starting at +10 m·s⁻¹ ✓ crossing the time axis at k = 1,02 s ✓ and continuing with the same constant negative gradient to −19,89 m·s⁻¹ at 3,05 s ✓; at 3,05 s the line jumps to a positive value smaller than 10 (≈ +13,28 m·s⁻¹, value not required) and decreases with the same gradient to zero at the top of the bounce ✓. Numerical values 10; 1,02; −19,89 shown.

Modelled on Nov 2025 Paper 1, Question 3 — same skills, new scenario.

Question 4

Object released from a moving carrierComprehension2 marksPaper 1

A delivery drone rises vertically at a constant velocity of 6 m·s⁻¹. A small parcel slips off the drone. Ignore air resistance.

Immediately after leaving the drone, the parcel ...

Hint

At the instant of release the parcel still has the drone's velocity; after that only gravity acts.

Worked answer

D. By inertia the parcel keeps the drone's upward velocity of 6 m·s⁻¹ at the instant of release; thereafter the only force is gravity, so it accelerates downwards at 9,8 m·s⁻² (it first rises, slowing down). A is the classic misconception that a released object starts from rest. B wrongly has it moving downwards immediately. C keeps the correct initial velocity but ignores gravity — free objects cannot have zero acceleration. Modelled on Nov 2025 Paper 1, Question 1.2 — same skills, new scenario.

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Grade 12 Physical Sciences: Vertical projectile motion

Question 1

Question 2

Question 3

Question 4

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