1phy easy question ...20marks

2012-06-14 3:03 am

a ball is dropped from a height h above the ground.each time it hits the ground, one half of the original kinetic energy is lost .where will ball rise above the ground after second impact?

回答 (1)

天同

2012-06-14 3:47 am

✔ 最佳答案

The kinetic energy of the ball K1 just before striking the ground
K1 = mgh
where m is the mass of the ball and g is the acceleration due to gravity
Hence, after the first impact, the kinetic energy possessed by the ball
= (K1)/2
= mgh/2
By conservation of energy, mg(h1) = mgh/2
where h1 is the height reached by the ball
h1 = h/2

Repeating the same calculation,
kinetic energy of the ball before striking the ground the 2nd time
K2 = mg(h1) = mgh/2
kinetic energy of ball after 2nd impact
= (K2)/2 = mgh/4
Hence, mgh/4 = mg(h2)
where h2 is the height reached by the ball after the 2nd impact
thus, h2 = h/4

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