Worksheet: Vertical Motion under Gravity

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In this worksheet, we will practice using Newton's laws of motion to model the vertical movement of an object in a straight line under gravity.

Q1:

A particle is projected vertically upwards at 7 m/s from a point 38.7 m above the ground. Find the maximum height the particle can reach. Consider the acceleration due to gravity to be 9.8 m/s2.

  • A 43.7 m
  • B 40.2 m
  • C 46.2 m
  • D 41.2 m

Q2:

A body was projected vertically upwards at 9.1 m/s. Determine the time taken to reach the maximum height, take 𝑔 = 9 . 8 / m s 2 .

  • A 1 3 2 8 s
  • B 1 3 7 s
  • C √ 9 1 7 s
  • D 1 3 1 4 s

Q3:

A ball is projected vertically upwards from the edge of a cliff, which is 186.2 m above the ground. If during the third second it covers a distance of 2.45 m upwards, find the time that the ball takes to reach the ground at the bottom of the cliff. Take the gravitational acceleration to be 9.8 m/s2.

Q4:

A body was projected vertically downwards from the top of a tower whose height is 80 m. Given that it covered 35.9 m during the 1st second of its motion, find the time taken to reach the ground rounded to the nearest two decimal places. Let the acceleration due to gravity 𝑔 = 9 . 8 / m s 2 .

Q5:

A body fell vertically from the top of a tower. It covered 86.73 m in the final second before hitting the ground. Determine the height of the tower rounding your answer to the nearest two decimal places. Let the acceleration due to gravity 𝑔 = 9 . 8 / m s 2 .

Q6:

A body was projected vertically downwards at 66.9 m/s from the top of a tower that is 457 metres high. Determine the distance that the body covered in the last second before it struck the ground. Take the acceleration due to gravity 𝑔 = 9 . 8 / m s 2 .

Q7:

Given that an object was projected vertically upwards at 619.92 km/h from the ground, what is the maximum height it can reach if the acceleration due to gravity is 9.8 m/s2?

Q8:

If a body, which was dropped from a building, took 3 seconds to reach the ground, find its average velocity as it fell. Let the acceleration due to gravity 𝑔 = 9 . 8 / m s 2 .

Q9:

A body was projected vertically upwards at 53.9 m/s. Given that at a certain time 𝑑 its height was 49 m, find all the possible values of 𝑑 . Take 𝑔 = 9 . 8 / m s 2 .

  • A 𝑑 = 0 . 8 4 s or 𝑑 = 1 1 . 8 4 s
  • B 𝑑 = 0 . 5 s or 𝑑 = 5 s
  • C 𝑑 = 1 0 s
  • D 𝑑 = 1 s or 𝑑 = 1 0 s

Q10:

A particle was projected vertically upwards from the ground. Given that the maximum height the particle reached was 62.5 m, find the velocity at which it was projected. Take the acceleration due to gravity 𝑔 = 9 . 8 / m s 2 .

Q11:

A body was projected vertically upwards at 14.7 m/s from a point 151 m above the ground. Find the position of the body 4 seconds after it was projected. Take .

  • A 131.4 m below the projection point
  • B 19.6 m above the projection point
  • C 53 m above the ground surface
  • D 131.4 m above the ground surface

Q12:

A particle was projected vertically upwards from the top of a tower. It was seen descending at the point of projection 8 seconds after it was projected. It reached the ground 5 seconds later. Find the height of the tower, 𝐻 , and the maximum height of the particle from the ground, 𝐻 m a x , take 𝑔 = 9 . 8 / m s  .

  • A 𝐻 = 4 4 1 m , 𝐻 = 3 9 6 . 9 m a x m
  • B 𝐻 = 3 1 8 . 5 m , 𝐻 = 3 9 2 m a x m
  • C 𝐻 = 4 4 1 m , 𝐻 = 3 9 2 m a x m
  • D 𝐻 = 3 1 8 . 5 m , 𝐻 = 3 9 6 . 9 m a x m

Q13:

A body was projected vertically upwards from the ground, and it took 157 seconds to return to the ground. Find the time 𝑑 1 for which the body was ascending and time 𝑑 2 for which it was descending.

  • A 𝑑 = 1 1 7 . 7 5 1 s , 𝑑 = 3 9 . 2 5 2 s
  • B 𝑑 = 3 9 . 2 5 1 s , 𝑑 = 1 1 7 . 7 5 2 s
  • C 𝑑 = 7 8 . 5 1 s , 𝑑 = 7 8 . 5 2 s

Q14:

A body was projected upwards at 34.3 m/s from the ground. It fell on the roof of a building 4.5 seconds after it was projected. Find, to the nearest two decimal places, the height of the building β„Ž 1 and the maximum height the body reached β„Ž 2 . Take the acceleration due to gravity 𝑔 = 9 . 8 / m s 2 .

  • A β„Ž = 2 7 . 5 6 1 m , β„Ž = 3 0 . 0 1 2 m
  • B β„Ž = 1 1 0 . 2 5 1 m , β„Ž = 1 2 0 . 0 5 2 m
  • C β„Ž = 1 3 . 7 8 1 m , β„Ž = 1 5 . 0 1 2 m
  • D β„Ž = 5 5 . 1 2 1 m , β„Ž = 6 0 . 0 2 2 m

Q15:

A particle was projected vertically upwards at 58.8 m/s from a point on the ground. 10.4 seconds later, another particle was projected from the same point at the same velocity. Find the time 𝑑 and the height β„Ž at which the two particles met. Take 𝑔 = 9 . 8 / m s 2 .

  • A 𝑑 = 2 6 . 8 s , β„Ž = 1 3 7 6 . 8 m
  • B 𝑑 = 1 6 . 4 s , β„Ž = 3 8 9 . 6 m
  • C 𝑑 = 1 2 . 5 3 s , β„Ž = 8 1 5 . 5 6 m
  • D 𝑑 = 1 1 . 2 s , β„Ž = 4 3 . 9 0 4 m

Q16:

A body was at rest at a height of 3.091 m above the ground. A wire lifted the body upwards causing it to accelerate at a rate of 1.89 m/s2. After moving for 2 seconds, the wire snapped. Find the velocity of the body 𝑣 right before the wire snapped and the maximum height that the body reached above the ground π‘₯ . Take the acceleration due to gravity 𝑔 = 9 . 8 / m s 2 .

  • A 𝑣 = 7 . 5 6 / m s , π‘₯ = 3 . 8 2 m
  • B 𝑣 = 1 . 8 9 / m s , π‘₯ = 7 . 5 6 m
  • C 𝑣 = 7 . 5 6 / m s , π‘₯ = 9 . 7 8 7 m
  • D 𝑣 = 3 . 7 8 / m s , π‘₯ = 7 . 6 m

Q17:

A body 𝐴 fell from a height of 425.4 m above the ground. At the same time, another body 𝐡 was projected vertically upwards at 70.9 m/s. Given that the two bodies crashed into one another, find the displacement of 𝐡 , from the point of its projection, when the two bodies collided. Take 𝑔 = 9 . 8 / m s 2 .

Q18:

Given that a ball, which was projected vertically upwards, covered 20.3 m during the sixth second of its motion, find the maximum height the ball reached. Take 𝑔 = 9 . 8 / m s 2 .

Q19:

A body was projected vertically downwards. It covered 23 m in the 2 n d second of its motion and 73 m in the 3 r d and 4 t h seconds. Find the velocity 𝑣 at which the body was projected and the acceleration due to gravity 𝑔 within the medium it was falling in.

  • A 𝑔 = 1 3 . 5 / m s 2 , 𝑣 = 1 6 . 6 7 / m s
  • B 𝑔 = 3 2 / m s 2 , 𝑣 = 5 9 . 5 / m s
  • C 𝑔 = 9 . 5 / m s 2 , 𝑣 = 8 / m s
  • D 𝑔 = 9 / m s 2 , 𝑣 = 9 . 5 / m s

Q20:

A stone was projected vertically upwards at 29.4 m/s from the front of a train of length 86 m. The train started moving at the moment the stone was projected. Given that the train accelerated at a rate of 4 m/s2, determine the distance between the rear end of the train and the point where the stone fell. Take the acceleration due to gravity 𝑔 = 9 . 8 / m s 2 .

Q21:

A body, at rest, fell from a height of 14.4 m onto a sandy surface. When it struck, it sank 46.8 cm into the sand. Find the total time taken from the moment the body began to fall until the moment it came to rest within the sand. Ignore any change in velocity due to the collision with the sandy ground. Take acceleration due to gravity to be 𝑔 = 9 . 8 / m s  .

Q22:

A body fell 5.62 m onto sandy ground. Given that it sank 56 cm into the sand before it came to rest, find the acceleration of the body as a result of it sinking into the sand. Take the acceleration due to gravity 𝑔 = 9 . 8 / m s 2 .

Q23:

An apple fell from a tree and took 0.8 s to reach the ground. Determine the original height of the apple from the ground, take 𝑔 = 9 . 8 / m s 2 .

Q24:

Determine the initial velocity of a ball that was projected vertically upwards, given that it covered a distance of 6.5 m during the third and fourth seconds of its motion and that the acceleration due to gravity 𝑔 = 9 . 8 / m s 2 .

Q25:

A ball, projected vertically upwards from a window, passed by the window again 5 seconds after the moment it was projected. Given that the gravitational acceleration is 𝑔 = 9 . 8 / m s 2 , determine the velocity at which the ball was projected.

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