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Question

A body is released from the top of a tower of height h. It takes v=12bt2+vo sec to reach the ground. Where will be the ball after time t2 sec.

A
At h/2 from the ground
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B
At h/4 from the ground
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C
Depends upon mass and volume of the body
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D
At 3h/4 from the ground
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Solution

The correct option is D At 3h/4 from the ground

let the body after time t2 be at x from the top,then

x=12gt24=gt28 (i)

h=12gt2 (ii)

Eliminate t from (i)and (ii),we get x=h4

Height of the body from the ground =hh4=3h4


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