WITH WHAT SPEED MUST THE BALL BE THROWM DOWN FOR IT TO BOUNCE 10m HIGHER THAN ITS ORIGINAL LEVEL NEGLECT ANY LOSS OF ENERGY IN STRIKING THE GROUND

Asked by shreya | 26th Oct, 2013, 08:52: PM

Expert Answer:

Suppose the ball is to be thrown from a height ‘h’ with a velocity ‘u’.

After bouncing on the ground it attains a maximum height (h+10).

Total energy at the initial point of throwing, E = mgh + ½ mu2

Total energy at the maximum height after bouncing back, E = mg (h+10)

By energy conservation,

mgh + ½ mu2 = mg(h+10)

=> ½ u2 = gh + 10g – gh

            = 10g

=> u = [20g]1/2 = (2 x 9.8)1/2 = (196)1/2 = 14 m/s

Answered by  | 28th Oct, 2013, 02:21: PM

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