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Kinematics and Projectile Motion MCQs — Physics (page 2)

79 multiple-choice questions on Kinematics and Projectile Motion from Physics, each with the correct answer and, where available, a teacher's explanation. Read through them to revise the chapter, then practise the same questions one by one to test yourself.

26. A ball is projected at 30 m/s at 45^circ (g = 9.8 m/s^2). Its horizontal range is:

  1. 91.9 m
  2. 91.8 m
  3. 45.9 m
  4. 184 m

Answer: B) 91.8 m

Explanation: R = u^2sin 2thetag = 91.8 m. SLO P-11-B-05 · Application

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27. A car starts with velocity 10 m/s and accelerates at 3 m/s^2 for 4 s. The distance covered is:

  1. 88 m
  2. 64 m
  3. 32 m
  4. 128 m

Answer: B) 64 m

Explanation: s = ut + 12at^2 = 10(4) + 12(3)(4)^2 = 64 m. SLO P-11-B-04 · Application

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28. A ball is projected at 40 m/s at 60^circ (g = 9.8 m/s^2). Its horizontal range is:

  1. 163 m
  2. 141 m
  3. 283 m
  4. 70.7 m

Answer: B) 141 m

Explanation: R = u^2sin 2thetag = 141 m. SLO P-11-B-05 · Application

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29. A ball is projected at 20 m/s at 30^circ (g = 9.8 m/s^2). Its maximum height is:

  1. 10.2 m
  2. 2.55 m
  3. 15.3 m
  4. 5.1 m

Answer: D) 5.1 m

Explanation: H = u^2sin^2theta2g = 5.1 m. SLO P-11-B-05 · Application

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30. Eliminating t between v = u + at and s = ( u + v2 )t gives:

  1. s = v^2a
  2. v = u^2 + 2as
  3. v^2 = u^2 - as
  4. v^2 = u^2 + 2as

Answer: D) v^2 = u^2 + 2as

Explanation: t = v - ua, so s = (v + u)(v - u)2a. SLO P-11-B-03 · Understanding

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31. A stone is dropped from 20 m (g = 9.8 m/s^2). Its speed on reaching the ground is:

  1. 19.8 m/s
  2. 9.9 m/s
  3. 39.6 m/s
  4. 14 m/s

Answer: A) 19.8 m/s

Explanation: v = 2gh = 2(9.8)(20) = 19.8 m/s. SLO P-11-B-04 · Application

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32. A ball rolls off a table 80 m high with horizontal speed 25 m/s (g = 9.8 m/s^2). It lands at a horizontal distance of:

  1. 50.5 m
  2. 101 m
  3. 143 m
  4. 202 m

Answer: B) 101 m

Explanation: Fall time t = 2h/g = 4.04 s; horizontal distance = vt = 101 m (horizontal velocity stays constant). SLO P-11-B-05 · Application

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33. At the highest point of its path, a projectile's velocity is:

  1. zero
  2. vertical
  3. horizontal, equal to ucostheta
  4. equal to u

Answer: C) horizontal, equal to ucostheta

Explanation: The vertical component is zero at the top. SLO P-11-B-05 · Application

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34. From the definition a = v - ut we obtain:

  1. v^2 = u^2 + 2as
  2. s = ut + 12at^2
  3. s = vt
  4. v = u + at

Answer: D) v = u + at

Explanation: Rearrange: v = u + at. SLO P-11-B-03 · Understanding

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35. A car starts with velocity 5 m/s and accelerates at 1.5 m/s^2 for 8 s. The distance covered is:

  1. 44 m
  2. 88 m
  3. 136 m
  4. 176 m

Answer: B) 88 m

Explanation: s = ut + 12at^2 = 5(8) + 12(1.5)(8)^2 = 88 m. SLO P-11-B-04 · Application

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36. A ball is projected at 20 m/s at 30^circ above the horizontal (g = 9.8 m/s^2). Its time of flight is:

  1. 2.04 s
  2. 4.08 s
  3. 1.02 s
  4. 3.53 s

Answer: A) 2.04 s

Explanation: T = 2usinthetag = 2(20)(0.5)9.8 = 2.04 s. SLO P-11-B-05 · Application

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37. With air resistance, the path of a projectile is:

  1. a straight line
  2. a circle
  3. a perfect parabola
  4. not symmetric — the descent is steeper than the ascent

Answer: D) not symmetric — the descent is steeper than the ascent

Explanation: The horizontal speed is smaller on the way down. SLO P-11-B-06 · Understanding

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38. A ball is thrown straight up at 29.4 m/s (g = 9.8 m/s^2). The maximum height reached is:

  1. 88.2 m
  2. 22 m
  3. 44.2 m
  4. 44.1 m

Answer: D) 44.1 m

Explanation: At the top v = 0: H = u^22g = 29.4^219.6 = 44.1 m. SLO P-11-B-04 · Application

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39. A ball is thrown straight up at 19.6 m/s. It returns to the thrower's hand after (g = 9.8 m/s^2):

  1. 5 s
  2. 2 s
  3. 4 s
  4. 8 s

Answer: C) 4 s

Explanation: Time up = u/g = 2 s; total time = 2u/g = 4 s. SLO P-11-B-04 · Application

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40. The area under a velocity–time graph gives:

  1. force
  2. velocity
  3. acceleration
  4. displacement

Answer: D) displacement

Explanation: Area = velocity × time. SLO P-11-B-03 · Understanding

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41. The slope of a velocity–time graph gives:

  1. jerk
  2. displacement
  3. speed
  4. acceleration

Answer: D) acceleration

Explanation: a = Delta vDelta t. SLO P-11-B-03 · Understanding

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42. With air resistance, the range of a projectile is:

  1. infinite
  2. exactly the same
  3. less than the ideal (no-air) range
  4. greater than the ideal range

Answer: C) less than the ideal (no-air) range

Explanation: Drag removes energy and slows the horizontal motion. SLO P-11-B-06 · Understanding

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43. A ball is projected at 30 m/s at 45^circ above the horizontal (g = 9.8 m/s^2). Its time of flight is:

  1. 2.16 s
  2. 4.34 s
  3. 8.66 s
  4. 4.33 s

Answer: D) 4.33 s

Explanation: T = 2usinthetag = 2(30)(0.707)9.8 = 4.33 s. SLO P-11-B-05 · Application

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44. A ball rolls off a table 45 m high with horizontal speed 15 m/s (g = 9.8 m/s^2). It lands at a horizontal distance of:

  1. 64.3 m
  2. 45.5 m
  3. 90.9 m
  4. 22.7 m

Answer: B) 45.5 m

Explanation: Fall time t = 2h/g = 3.03 s; horizontal distance = vt = 45.5 m (horizontal velocity stays constant). SLO P-11-B-05 · Application

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45. A ball is projected at 25 m/s at 53.13^circ (g = 9.8 m/s^2). Its maximum height is:

  1. 11.5 m
  2. 10.2 m
  3. 20.4 m
  4. 40.8 m

Answer: C) 20.4 m

Explanation: H = u^2sin^2theta2g = 20.4 m. SLO P-11-B-05 · Application

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46. At the highest point, a projectile's acceleration is:

  1. g upward
  2. zero
  3. horizontal
  4. g downward

Answer: D) g downward

Explanation: Gravity acts throughout the flight. SLO P-11-B-05 · Application

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47. In s = ut + 12at^2, the term ut represents:

  1. the distance covered if there were no acceleration
  2. the final velocity
  3. the extra distance due to acceleration
  4. the average velocity

Answer: A) the distance covered if there were no acceleration

Explanation: It is the rectangle part of the v–t area. SLO P-11-B-03 · Understanding

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48. A body's velocity changes from 10 m/s to 30 m/s over 100 m. Its acceleration is:

  1. 4 m/s^2
  2. -4 m/s^2
  3. 2 m/s^2
  4. 8 m/s^2

Answer: A) 4 m/s^2

Explanation: a = v^2 - u^22s = 900 - 1002(100) = 4 m/s^2. SLO P-11-B-04 · Application

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49. A stone is dropped from a height of 45 m (g = 9.8 m/s^2, no air resistance). It reaches the ground after:

  1. 1.52 s
  2. 3.03 s
  3. 4.29 s
  4. 6.06 s

Answer: B) 3.03 s

Explanation: h = 12gt^2 Rightarrow t = 2hg = 2(45)9.8 = 3.03 s. SLO P-11-B-04 · Application

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50. A car starts with velocity 10 m/s and accelerates at 3 m/s^2 for 4 s. Its final velocity is:

  1. 44 m/s
  2. 22 m/s
  3. 11 m/s
  4. 33 m/s

Answer: B) 22 m/s

Explanation: v = u + at = 10 + (3)(4) = 22 m/s. SLO P-11-B-04 · Application

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