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⁻¹. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take g = 10 m s⁻² ) :

64 m

$ \text{ Time taken by ball to come back to bridge is } T_1 = \frac{2u}{g} = \frac{2\times 4}{10} = 0.8 s$ 

$\text{Hence time taken by ball to reach river from bridge is }4- 0.8 = 3.2 s $

$\text{ Let height of bridge is h , for motion below the bridge u = -4 m/s , g = -10 m/s^2}$

$\Rightarrow  h = ut + \frac{1}{2} at^2 $

$ -h = -4t - 5t^2 $

$\Rightarrow h = 4\times 3.2 + 5 \times 3.2^2 = 64m$