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Kinematics and Projectile Motion MCQs — Physics (page 3)

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.

51. The range of a projectile on level ground is maximum when it is launched at:

  1. 30^circ
  2. 45^circ
  3. 90^circ
  4. 60^circ

Answer: B) 45^circ

Explanation: sin 2theta is largest (1) when 2theta = 90^circ. SLO P-11-B-05 · Application

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52. A car starts with velocity 0 m/s and accelerates at 4 m/s^2 for 3 s. The distance covered is:

  1. 9 m
  2. 18 m
  3. 19 m
  4. 36 m

Answer: B) 18 m

Explanation: s = ut + 12at^2 = 0(3) + 12(4)(3)^2 = 18 m. SLO P-11-B-04 · Application

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53. A ball is projected at 50 m/s at 30^circ above the horizontal (g = 9.8 m/s^2). Its time of flight is:

  1. 8.84 s
  2. 2.55 s
  3. 10.2 s
  4. 5.1 s

Answer: D) 5.1 s

Explanation: T = 2usinthetag = 2(50)(0.5)9.8 = 5.1 s. SLO P-11-B-05 · Application

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54. A projectile launched at 20^circ has the same range as one launched (same speed) at:

  1. 80^circ
  2. 70^circ
  3. 25^circ
  4. 40^circ

Answer: B) 70^circ

Explanation: Angles that add to 90^circ give the same range. SLO P-11-B-05 · Application

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55. The equations of motion apply only when:

  1. the acceleration is uniform (constant)
  2. the body moves in a circle
  3. the acceleration is changing
  4. the velocity is constant

Answer: A) the acceleration is uniform (constant)

Explanation: They are derived assuming constant a. SLO P-11-B-03 · Understanding

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56. Two projectiles launched with the same speed at 30^circ and 60^circ have:

  1. the same time of flight
  2. different ranges
  3. the same maximum height
  4. the same range

Answer: D) the same range

Explanation: Complementary angles give equal sin 2theta. SLO P-11-B-05 · Application

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57. The slope of a displacement–time graph gives:

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

Answer: B) velocity

Explanation: v = Delta sDelta t. SLO P-11-B-03 · Understanding

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58. A ball is projected at 40 m/s at 60^circ (g = 9.8 m/s^2). Its maximum height is:

  1. 122 m
  2. 20.4 m
  3. 61.2 m
  4. 30.6 m

Answer: C) 61.2 m

Explanation: H = u^2sin^2theta2g = 61.2 m. SLO P-11-B-05 · Application

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59. Air resistance acts:

  1. perpendicular to the velocity
  2. always upward
  3. always downward
  4. opposite to the direction of the velocity

Answer: D) opposite to the direction of the velocity

Explanation: Drag opposes motion through the air. SLO P-11-B-06 · Understanding

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60. With air resistance, the angle of launch for maximum range is:

  1. exactly 45^circ
  2. less than 45^circ
  3. more than 45^circ
  4. 90^circ

Answer: B) less than 45^circ

Explanation: A lower angle reduces the time spent in the air where drag acts. SLO P-11-B-06 · Understanding

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61. The ideal parabolic path is a good model for:

  1. a badminton shuttle
  2. a parachute
  3. slow, dense objects over short distances
  4. a falling leaf

Answer: C) slow, dense objects over short distances

Explanation: Drag is then small compared with gravity. SLO P-11-B-06 · Understanding

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62. The area of the trapezium splits into a rectangle ut and a triangle of area:

  1. 12at^2
  2. at^2
  3. ut^2
  4. 12at

Answer: A) 12at^2

Explanation: Triangle: 12 t (v - u) = 12t(at). SLO P-11-B-03 · Understanding

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63. A body's velocity changes from 0 m/s to 20 m/s over 50 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 = 400 - 02(50) = 4 m/s^2. SLO P-11-B-04 · Application

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64. A ball is projected at 25 m/s at 53.13^circ above the horizontal (g = 9.8 m/s^2). Its time of flight is:

  1. 2.04 s
  2. 3.06 s
  3. 4.08 s
  4. 8.16 s

Answer: C) 4.08 s

Explanation: T = 2usinthetag = 2(25)(0.8)9.8 = 4.08 s. SLO P-11-B-05 · Application

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65. A stone is dropped from a height of 20 m (g = 9.8 m/s^2, no air resistance). It reaches the ground after:

  1. 1.01 s
  2. 2.02 s
  3. 4.04 s
  4. 2.86 s

Answer: B) 2.02 s

Explanation: h = 12gt^2 Rightarrow t = 2hg = 2(20)9.8 = 2.02 s. SLO P-11-B-04 · Application

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66. Compared with the ideal case, the time of flight with air resistance is usually:

  1. exactly the same up and down
  2. zero
  3. doubled
  4. slightly different — the descent takes longer than the ascent

Answer: D) slightly different — the descent takes longer than the ascent

Explanation: The ball rises faster than it falls back. SLO P-11-B-06 · Understanding

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67. A ball is thrown straight up at 19.6 m/s (g = 9.8 m/s^2). The maximum height reached is:

  1. 19.6 m
  2. 39.2 m
  3. 9.8 m
  4. 19.7 m

Answer: A) 19.6 m

Explanation: At the top v = 0: H = u^22g = 19.6^219.6 = 19.6 m. SLO P-11-B-04 · Application

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68. For a projectile (no air resistance), the horizontal component of velocity:

  1. increases
  2. stays constant
  3. changes direction
  4. decreases to zero at the top

Answer: B) stays constant

Explanation: No horizontal force acts. SLO P-11-B-05 · Application

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69. A ball is projected at 40 m/s at 60^circ above the horizontal (g = 9.8 m/s^2). Its time of flight is:

  1. 3.53 s
  2. 7.07 s
  3. 14.1 s
  4. 4.08 s

Answer: B) 7.07 s

Explanation: T = 2usinthetag = 2(40)(0.866)9.8 = 7.07 s. SLO P-11-B-05 · Application

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70. A stone is dropped from 80 m (g = 9.8 m/s^2). Its speed on reaching the ground is:

  1. 28 m/s
  2. 19.8 m/s
  3. 79.2 m/s
  4. 39.6 m/s

Answer: D) 39.6 m/s

Explanation: v = 2gh = 2(9.8)(80) = 39.6 m/s. SLO P-11-B-04 · Application

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71. A straight line sloping downward on a v–t graph shows:

  1. increasing acceleration
  2. uniform acceleration
  3. uniform deceleration
  4. constant velocity

Answer: C) uniform deceleration

Explanation: Negative constant slope. SLO P-11-B-03 · Understanding

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72. On a v–t graph of uniform acceleration from u to v, the area is a trapezium of area:

  1. 12(u + v)t
  2. (v - u)t
  3. uvt
  4. 12(v - u)t

Answer: A) 12(u + v)t

Explanation: Area of trapezium = average of parallel sides × width. SLO P-11-B-03 · Understanding

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73. A car starts with velocity 20 m/s and accelerates at -2 m/s^2 for 6 s. The distance covered is:

  1. 84 m
  2. 48 m
  3. 168 m
  4. 42 m

Answer: A) 84 m

Explanation: s = ut + 12at^2 = 20(6) + 12(-2)(6)^2 = 84 m. SLO P-11-B-04 · Application

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74. A ball is projected at 50 m/s at 30^circ (g = 9.8 m/s^2). Its horizontal range is:

  1. 221 m
  2. 442 m
  3. 255 m
  4. 110 m

Answer: A) 221 m

Explanation: R = u^2sin 2thetag = 221 m. SLO P-11-B-05 · Application

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75. A horizontal straight line on a v–t graph shows:

  1. constant velocity (zero acceleration)
  2. constant acceleration
  3. the body at rest always
  4. decreasing speed

Answer: A) constant velocity (zero acceleration)

Explanation: Zero slope means zero acceleration. SLO P-11-B-03 · Understanding

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