Gravitation- numerical probelm, Physics

A ball dropped from a balloon at rest clears a tower  81m high during the last quater second of its journey. Find the height of the balloon and the velocity of the body when it reaches the ground.

Solution) The last quarter of the jouney menas 1/4th of the total time it requires in its total journey.

Let total time be 4t, u = 0.

So, v at 3t = 3gt, and at 4t = 4gt.

by v² = u² + 2as,

16g²t² = 9g²t² + 2g(81)

7g²t² = 162g

7gt2 = 162

t² = 162 / 7g

t = 1.5 sec.

but time is 4t = 6 sec.

So velocity on reaching = 4gt = 58.8 m/s

With this velocity, height of fall = 2t(58.8) = 705.6m

Posted Date: 3/11/2013 7:13:32 AM | Location : United States

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