IB · Physics SL & HL · A Space, time & motion

A.1 Kinematics: IB Physics questions with answers

Free A.1 Kinematics questions for IB Physics (SL & HL), part of A Space, time & motion. Each question has the answer and the marking points an examiner would look for.

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A.1 Kinematics practice questions

Question 1Multiple choice · 1 markA* challenge

An object starts from rest. Its velocity increases uniformly to 4 m s−1 in 3 s and then decreases uniformly to zero after a further 3 s. It then accelerates uniformly in the opposite direction, reaching 2 m s−1 after 2 s, and slows uniformly to rest after a further 2 s.
What is the magnitude of the displacement of the object from its starting point?

  1. A 8.0 m
  2. B 9.0 m
  3. C 12 m
  4. D 16 m
Show answer

Answer: A

Question 2Multiple choice · 1 markA* challenge

A ball is projected at an angle to the horizontal. Air resistance is NOT negligible. Compared with the motion without air resistance, which statements are correct?
I.   The horizontal range is smaller.
II.   The time to fall from the maximum height is greater than the time to rise.
III.   The acceleration at the highest point is exactly g.

  1. A I and II only
  2. B I and III only
  3. C II and III only
  4. D I, II and III
Show answer

Answer: A

Question 3Multiple choice · 1 mark

A projectile is launched at an angle above the horizontal. Air resistance is negligible.
What are the velocity and the acceleration of the projectile at its highest point?

velocityacceleration
Ahorizontal, non-zerozero
Bzerog, vertically downwards
Czerozero
Dhorizontal, non-zerog, vertically downwards
Show answer

Answer: D

Question 4Structured · 11 marks

A stone is thrown from the top of a vertical cliff 20 m above the sea. Its initial velocity is 20 m s−1 at 40° above the horizontal. Air resistance is negligible.

(a)

Calculate the maximum height of the stone above the sea.

[2]
(b)

Show that the stone takes about 3.72 s to reach the sea.

[2]
(c)

Determine the horizontal distance from the foot of the cliff to where the stone lands.

[1]
(d)

Determine the speed of the stone as it hits the sea and the angle its velocity makes with the horizontal.

[3]
(e)

Air resistance is now taken into account. State and explain the effect on the horizontal distance and on the angle at which the stone hits the sea.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)28.4 m
vertical component 12.9 m s−1; rise = vy22g = 8.42 m ✓
height = 28.4 m ✓
2
(b)3.72 s
−20 = 12.9t − 4.91t2 ✓
solving: t = 3.72 s ✓
2
(c)57.0 m
15.3 × 3.72 = 57.0 m ✓
1
(d)28.1 m s−1 at 57.0° below horizontal
vy = 12.9 − 9.81(3.72) = −23.6 m s−1 ✓
speed = 28.1 m s−1 ✓
angle = 57.0° below the horizontal ✓
3
(e)distance less, angle steeper
the horizontal component of velocity decreases throughout, so the distance is smaller ✓
the horizontal velocity at impact is reduced more than the vertical ✓
so the stone hits the sea at a steeper angle ✓
3
Question 5Structured · 7 marks

A ball is thrown horizontally at 12 m s−1 from the top of a cliff 45 m high. Air resistance is negligible.

(a)

Calculate the time taken for the ball to reach the ground.

[2]
(b)

Calculate the horizontal distance travelled.

[1]
(c)

Calculate the speed of the ball as it hits the ground.

[2]
(d)

Air resistance is now taken into account. Explain the effect on the horizontal distance travelled.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)3.0 s
45 = ½ × 9.8 × t2 ✓
t = 3.03 s ✓
2
(b)36 m
12 × 3.03 = 36.4 m ✓
1
(c)32 m s−1
vy = 9.8 × 3.03 = 29.7 m s−1 ✓
v = √(122 + 29.72) = 32.0 m s−1 ✓
2
(d)distance decreases
horizontal velocity decreases during the flight ✓
so the horizontal distance is smaller ✓
2
Question 6Structured · 10 marksA* challenge

A projectile is launched from level ground at 20 m s−1 at 35° to the horizontal. Air resistance is negligible.

(a)

Calculate the horizontal and vertical components of the initial velocity.

[2]
(b)

Calculate the maximum height reached.

[2]
(c)

Calculate the horizontal range.

[3]
(d)

A wall 4.0 m high stands 30 m from the launch point. Determine whether the projectile clears the wall.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)16.4 and 11.5 m s−1
vx = 20cos35° = 16.4 m s−1 ✓
vy = 20sin35° = 11.5 m s−1 ✓
2
(b)6.7 m
h = vy2/2g ✓
= 11.472/19.6 = 6.71 m ✓
2
(c)38 m
t = 2 × 11.47/9.8 ✓
= 2.34 s ✓
R = 16.38 × 2.34 = 38.3 m ✓
3
(d)yes, 4.6 m
t = 30/16.38 = 1.83 s ✓
y = 11.47 × 1.83 − 4.9 × 1.832 = 4.6 m ✓
4.6 m > 4.0 m so it clears the wall ✓
3

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in