A stone is thrown vertically upward with an initial speed $ u $ from the top of a tower, reaches the ground with a speed $ 3 u $. The height of the tower is:

Question

A stone is thrown vertically upward with an initial speed $ u $ from the top of a tower, reaches the ground with a speed $ 3 u $. The height of the tower is:, $ \frac{3 u^{2}}{g} $, $ \frac{4 u^{2}}{g} $, $ \frac{6 u^{2}}{g} $, $ \frac{9 u^{2}}{g} $

MARKS App · Accepted Answer

Let the height of the tower be h . Taking + v e x - axis along vertically upward direction, we have, u = u a = - g v = - 3 u s = A C = - h As acceleration remains constant. Now using v 2 = u 2 + 2 a s , we get ⇒ - 3 u 2 = u 2 + 2 - g - h ⇒ 9 u 2 = u 2 + 2 g h ⇒ 2 g h = 8 u 2 ⇒ g h = 4 u 2 ⇒ h = 4 u 2 g .

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