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A body is released from the top of a towerof height h. It takes sec to reach theground. Where will be the ball after timet12 sec(A) At h/2 from the groundB) At h/4 from the ground(C) Depends upon mass and volume of the) At 3h/4 from the ground |
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Answer» from the equation of motion we know thah =1/2(g)t² for t = t/2h' = 1/2*(g)(t/2)² = 1/2*g*t²/4 therefore h' = 1/4*(1/2*g*t²) = 1/4*h so, from the distance it is at h/4 from the ground it is at h-h/4 = 3h/4 distance. |
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