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Question

A ball is projected horizontally with a velocity of 4ms from the top of a building 19.6m high. The distance of the point from the bottom of the building where the ball will hit the ground is g=9.8ms2


  1. 8m

  2. 6.6m

  3. 20.6m

  4. 14.6m

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Solution

The correct option is A

8m


Step 1. Given data:

  1. Height of the building = 19.6m
  2. Since the ball is projected horizontally, the component of velocity in the horizontal direction Vx=4ms,
  3. And its initial vertical velocity, the component of velocity in the vertical direction Vy=0
  4. So the time required to strike the ground will be free-fall time.
  5. Acceleration due to gravity = 9.8ms2

Step 2. Formula used:

  1. Time of flight, t=2hg, Where h=height of the building, g=acceleration due to gravity.
  2. Also Sx=ux.t+12at2Sx=ux.t, Where Sx=horizontal distance in the x-direction.,a=0 for the horizontal direction.

Step 3. Calculations:

Putting all the known values in the above equations, we get

t=2×19.69.8=2s

Hence, Sx=ux.tSx=4×2=8m,

Thus, the distance of the point from the bottom of the building where the ball will hit the ground is 8m.

Hence, option A is the correct option.


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