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NEET PHYSICSMedium

A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s−14 \text{ m s}^{-1}. The ball strikes the water surface after 4 s4 \text{ s}. The height of bridge above water surface is (Take g=10 m s−2g = 10 \text{ m s}^{-2})

A

56 m

B

60 m

C

64 m

D

68 m

Step-by-Step Solution

Using the equation of motion s=ut+12at2s = ut + \frac{1}{2}at^2. Taking the upward direction as positive, u=4 m s−1u = 4 \text{ m s}^{-1}, a=−g=−10 m s−2a = -g = -10 \text{ m s}^{-2}, t=4 st = 4 \text{ s}. The displacement s=−hs = -h (where hh is the height of the bridge). So, −h=(4)(4)+12(−10)(4)2=16−80=−64-h = (4)(4) + \frac{1}{2}(-10)(4)^2 = 16 - 80 = -64. Thus, h=64 mh = 64 \text{ m}.

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