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A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 ms^–1. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take g = 10 m s^–2):

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68 m

56 m

60 m

64 m

answer is D.

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Detailed Solution

Using equation of motion, considering the upward direction as positive
Total displacement = –H (Height of bridge ) Time taken = 4s

$\Rightarrow - H = ut + \frac{1}{2} (- g) t^{2}$

Substituting the value of t = 4s

$\begin{aligned} \Rightarrow - H = 4 (4) + \frac{1}{2} (- 10) (4)^{2} \\ \Rightarrow H = 64 m \end{aligned}$

$∴ Height of bridge = 64 m$

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