Q.

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 the bridge above water surface is (Take g=10 ms-2)          [2023]

1 64 m  
2 68 m  
3 56 m  
4 60 m  

Ans.

(1)

Given, u=4 ms-1, v=0, g=10 ms-2

Time taken by ball to reach its maximum height,

     v=u+at

⇒0=4+(-10)t or 4=10t

      t=0.4 sec.

The height of bridge above water surface is

S=ut+12at2⇒S=(0)t+12(10)(3.6)2

S=64 m