Question

particle of mass 1kg dropped from height 20 metre and after striking the ground it rebounds to height 5 metre if girls contact time with ground is .01 second then calculate a force acting on the ball​

Answer 1

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mass, m = 1kg

height, s = 20m

initial velocity of ball, u = 0m/s

acceleration, a = 10m/s

Using, v

= 0 + 2×10 ×20

= 400

v = 20m/s

Now, let us take upward direction as positive and downward direction as negative.

Initial momentum of the ball (before striking the ground) = m×v

= 1×(-20) (velocity is downward, he

= -20kgm/s

Now, when the ball strikes and rebounds, its velocity becomes +20m/s (after striking, it moves upward, hence positive velocity)

Final momentum (after striking the ground) = m×v

= 1×(+20)

= +20kgm/s

Change in momentum = Final momentum - Initial momentum

= 20kgm/s - (-20kgm/s)

= 40kgm/s in upward direction

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Answer 2

Explanation:

hence force on ball is 990 Newton

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