Kinematics Practice Questions (Set 1)

Exam-style A Level kinematics questions on graphs, projectile components, gravitational energy and air resistance, with worked solution checks.

GCE A-Level H2 Physics 2027

Back to topic: Motion in a Gravitational Field: projectiles, energy and drag

Learning goals

Work through the questions and use the feedback to plan revision.

Learning objectives

Interpret position, displacement, velocity and acceleration using equations and graphs.
Derive the uniformly accelerated motion equations from the definitions of velocity and acceleration.
Derive and apply uniformly accelerated motion equations with a stated sign convention.
Relate weight and gravitational potential energy changes in a uniform gravitational field.
Analyse projectile motion by separating perpendicular components.
Explain falling motion with air resistance using forces, energy and terminal velocity.

Attempt each question first, then reveal the solution check. Write separate horizontal and vertical equations, define upward as positive unless another choice is clearer, and keep unrounded values until the final answer.

Use the value of g stated in the question. If none is given, take g = 9.81 m s⁻².

Questions

1) Horizontal launch: rifle and target

A rifle is aimed horizontally at a target 30 m away. The bullet hits the target 1.9 cm below the aiming point. Calculate:

(a) the bullet’s time of flight
(b) its speed as it emerges from the rifle

Click to show/hide answer

Take downward as positive. Vertically, s = 0.019 m, u_y = 0 and a = g:

t = square root of (2s/g) = square root of (2(0.019)/9.81) = 0.0622 s

Horizontally, vₓ = x/t = 30/0.0622 = 4.82 × 10² m s⁻¹.

(a) 62 ms
(b) 4.8 × 10² m s⁻¹

2) Skier leaving a slope (landing below)

A skier leaves a slope at 20° above the horizontal with a speed of 10 m s⁻¹. The horizontal landing surface is 3 m below the slope.

(a) At what horizontal distance from the launching point does she land?
(b) At what impact velocity will she land?

Click to show/hide answer

Take upwards as positive. uₓ = 10 cos 20° = 9.40 m s⁻¹ and u_y = 10 sin 20° = 3.42 m s⁻¹.

Solve -3.00 = 3.42t-(1/2)(9.81)t² to get t = 1.205 s. Hence x = uₓt = 11.3 m.

At impact, vₓ = 9.40 m s⁻¹ and v_y = 3.42-9.81(1.205) = -8.40 m s⁻¹. Therefore v = 12.6 m s⁻¹ at tan⁻¹ (8.40/9.40) = 41.8° below the horizontal.

(a) 11.3 m
(b) 12.6 m s⁻¹, 41.8° below the horizontal

3) Tabletop launch

A small ball rolls horizontally off the edge of a tabletop that is 1.2 m high. It strikes the floor at a point 1.52 m horizontally away from the edge of the table. Calculate:

(a) the time for the ball to remain in the air
(b) the speed of the ball at the instant it leaves the table

Click to show/hide answer

The vertical fall fixes the time:

t = square root of (2(1.2)/9.81) = 0.495 s

The launch speed is horizontal, so uₓ = 1.52/0.495 = 3.07 m s⁻¹.

(a) 0.495 s
(b) 3.07 m s⁻¹

4) Ball thrown at a wall

You throw a ball with a speed of 25.3 m s⁻¹ at an angle of 42.0° above the horizontal directly towards a wall. The wall is 21.8 m from the release point of the ball.

(a) How long is the ball in the air before it hits the wall?
(b) How far above the release point does the ball hit the wall?
(c) What are the horizontal and vertical components of its velocity as it hits the wall?

Click to show/hide answer

The horizontal velocity is vₓ = 25.3 cos 42.0° = 18.8 m s⁻¹, so

t = 21.8/18.8 = 1.16 s.

Using u_y = 25.3 sin 42.0° = 16.9 m s⁻¹ gives y = u_yt-(1/2)gt² = 13.0 m and v_y = u_y-gt = +5.55 m s⁻¹.

(a) 1.16 s
(b) 13.0 m
(c) vₓ = 18.8 m s⁻¹, v_y = +5.55 m s⁻¹ (still upwards)

5) Long jump range (optional)

In the 1991 World Track and Field Championships in Tokyo, Mike Powell jumped 8.95 m, breaking the 23-year long jump record by a full 5 cm. Assuming that Powell’s speed on takeoff was 9.5 m s⁻¹ (about equal to that of a sprinter) and that g is 9.80 m s⁻² in Tokyo, how much less was Powell’s horizontal range than the maximum possible range (neglect air resistance) for a particle launched at the same speed?

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For equal launch and landing heights, R = u² sin 2θ/g. The maximum occurs when sin 2θ = 1, so

Rₘₐₓ = 9.5²/9.80 = 9.21 m.

The difference is 9.21-8.95 = 0.26 m.

6) Baseball clearing a wall

A home run is hit in such a way that the baseball just clears a wall 21 m high, located 130 m from the home plate. The bat hits the ball when the ball is 1 m above the ground. The ball is hit at an angle of 35° to the horizontal, and air resistance is negligible. Find:

(a) the initial speed of the ball
(b) the time it takes the ball to reach the wall
(c) the velocity of the ball when it reaches the wall (state components, or speed + direction)

Click to show/hide answer

Relative to the launch point, the wall top is y = 21-1 = 20 m. Substitute x = 130 m and θ = 35° into

y = x tan θ-gx²/(2u² cos² θ)

to obtain u = 41.7 m s⁻¹. Then uₓ = 34.2 m s⁻¹ and t = x/uₓ = 3.81 s. At the wall, v_y = u sin 35°-gt = -13.4 m s⁻¹.

(a) 41.7 m s⁻¹
(b) 3.81 s
(c) vₓ = 34.2 m s⁻¹, v_y = -13.4 m s⁻¹; equivalently 36.7 m s⁻¹ at 21.4° below horizontal

7) Rock launched from a cliff

A rock is projected 20° above the horizontal at 5.0 m s⁻¹ from a cliff 50 m above the ground. Find:

(a) the horizontal distance from the base of the cliff where the rock lands
(b) the impact velocity

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uₓ = 5.0 cos 20° = 4.70 m s⁻¹ and u_y = 5.0 sin 20° = 1.71 m s⁻¹. Solve

-50 = 1.71t-(1/2)(9.81)t²

to get t = 3.37 s and x = uₓt = 15.8 m. At impact, v_y = 1.71-9.81(3.37) = -31.4 m s⁻¹.

(a) 15.8 m
(b) vₓ = 4.70 m s⁻¹, v_y = -31.4 m s⁻¹; speed 31.7 m s⁻¹ at 81.5° below horizontal

8) Player running to meet a kicked ball

A soccer ball is kicked from the ground with an initial speed of 19.5 m s⁻¹ at an upward angle of 45°. A player 55 m away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? Neglect air resistance.

Click to show/hide answer

For a 45° launch returning to the same height,

T = (2u sin 45°)/g = 2.81 s, R = u²/g = 38.8 m.

The player must cover 55.0-38.8 = 16.2 m in 2.81 s, so the required average speed is 5.78 m s⁻¹ towards the launch point.

9) Ball rolling off a stairway

A ball rolls horizontally off the top of a stairway with a speed of 1.52 m s⁻¹. The steps are 20.3 cm high and 20.3 cm wide. Which step does the ball hit first?

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Let each step have height and width d = 0.203 m. The trajectory measured downwards from the top edge is

y = (g/2uₓ²)x².

At the outer edge of step n, x = nd; the ball has fallen yₙ = g(nd)²/(2uₓ²). Checking successive edges gives y₁ = 0.086 m < d and y₂ = 0.344 m < 2d, while the trajectory intersects the horizontal surface of the third step before its outer edge. Therefore it first hits the 3rd step.

10) Tennis serve and the net

During a tennis match, a player serves the ball at 23.6 m s⁻¹, with the centre of the ball leaving the racquet horizontally 2.37 m above the court surface. The net is 12 m away and 0.90 m high.

When the ball reaches the net:

(a) Does the ball clear it?
(b) What is the distance between the centre of the ball and the top of the net?

Suppose instead the ball is served as before but now it leaves the racquet at 5° below the horizontal. When the ball reaches the net:

(c) Does the ball clear it?
(d) What now is the distance between the centre of the ball and the top of the net?

Click to show/hide answer

The time to the net is t = 12/23.6 = 0.508 s.

For the horizontal serve, the ball height is 2.37-(1/2)(9.81)t² = 1.10 m, which is 0.20 m above the net.

For the 5° downward serve, u_y = -23.6 sin 5° = -2.06 m s⁻¹ and uₓ = 23.6 cos 5°, so t = 12/uₓ = 0.510 s. The centre is then about 0.05 m above the court, or 0.85 m below the net top.

(a) Yes
(b) 0.20 m
(c) No
(d) 0.85 m

11) Projectile launched from ground and from a building

(a) A body is projected from ground level at an angle of 50° above the horizontal with an initial speed of 40 m s⁻¹. How long will it be before it hits the ground?
(b) A body is projected at an angle of 30° below the horizontal from the top of a building 170 m high. Its initial speed is 40 m s⁻¹. How long will it take before striking the ground? What is the velocity of the body just before it touches the ground?

Click to show/hide answer

For (a), equal launch and landing heights give

T = (2(40 sin 50°))/9.81 = 6.25 s.

For (b), uₓ = 40 cos 30° = 34.6 m s⁻¹ and u_y = -40 sin 30° = -20.0 m s⁻¹. Solve -170 = -20.0t-(1/2)(9.81)t² to obtain t = 4.19 s. Then v_y = -20.0-9.81(4.19) = -61.1 m s⁻¹.

(a) 6.2 s (to 2 s.f.)
(b) 4.19 s; vₓ = 34.6 m s⁻¹, v_y = -61.1 m s⁻¹ (speed 70.2 m s⁻¹, 60.5° below horizontal)

12) Range and take-off speed

(a) A football is kicked at a speed of 20 m s⁻¹ at a launch angle of 45°. Find its range assuming negligible air resistance.
(b) A stunt driver leaves a ramp at an angle of 30° and lands at the same height after travelling a horizontal distance of 36 m. What was his take-off speed?

Click to show/hide answer

For equal launch and landing heights, R = u² sin 2θ/g.

For (a), R = 20² sin 90°/9.81 = 40.8 m.

For (b), rearrange to u = square root of (Rg/sin 2θ):

u = square root of (36(9.81)/(sin 60°)) = 20.2 m s⁻¹.

(a) 40.8 m
(b) 20.2 m s⁻¹

13) Weight and gravitational potential energy

A 0.50 kg ball descends through a vertical height of 12.0 m near Earth’s surface. Take g = 9.81 N kg⁻¹ and neglect air resistance.

(a) Calculate the ball’s weight.
(b) Calculate the change in gravitational potential energy of the ball–Earth system.
(c) State the corresponding change in kinetic energy.

Click to show/hide answer

Weight is the gravitational force:

W = mg = (0.50)(9.81) = 4.91 N.

Take upwards as positive, so Δ h = -12.0 m:

Δ Eₚ = mgΔ h = (0.50)(9.81)(-12.0) = -58.9 J.

With no resistive loss, mechanical energy is conserved. The kinetic energy therefore increases by 58.9 J.

(a) 4.91 N downwards
(b) Δ Eₚ = -58.9 J
(c) Δ Eₖ = +58.9 J

14) Falling with air resistance

A ball is released from rest and falls far enough to reach terminal velocity. Describe how its resultant force, acceleration and energy transfers change from release until it is moving at terminal velocity.

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Just after release, drag is zero while weight acts downwards, so the resultant force and acceleration are greatest and downward.
As speed rises, upward drag increases. The downward resultant force and acceleration decrease.
At terminal velocity, drag equals weight. Resultant force and acceleration are zero, while velocity is constant and non-zero.
Before terminal velocity, decreasing gravitational potential energy is transferred to kinetic energy and thermal energy of the ball and surroundings.
At terminal velocity, kinetic energy stays constant; the continuing loss of gravitational potential energy is transferred to thermal energy.

Do not say that “no acceleration” means “no motion”, or that the falling ball’s speed decreases throughout.

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Course and syllabus information

Course: GCE A-Level H2 Physics
Edition: GCE A-Level H2 Physics 2027