Q: A ball is thrown upwards with a speed V from a height ‘h’ above the ground. The time taken by the ball to hit the ground is

(a) $ \displaystyle \sqrt{\frac{2 h}{g}} $

(b) $ \displaystyle \sqrt{\frac{8 h}{g}} $

(c) $ \displaystyle \frac{\sqrt{V^2 + 2 g h}}{g} + \frac{V}{g} $

(d) $ \displaystyle \sqrt{\frac{2 h}{g}} + \frac{V}{g} $

Ans:(c)

Sol : Time taken to reach highest point from the height h is obtained from

v = u + at

0 = V – gt,

t = V/g …(i)

Now , Height attained is obtained from

$ \displaystyle v^2 = u^2 + 2 a s $

$ \displaystyle 0 = V^2 – 2 g s $

$ \displaystyle s = \frac{V^2}{2 g} $

Total height for downward journey is

$ \displaystyle = \frac{V^2}{2 g} + h $

Time taken to hit the ground is obtained from

$\displaystyle s = ut + \frac{1}{2}a t^2 $

$ \displaystyle \frac{V^2}{2 g} + h = 0 + \frac{1}{2}g t^2 $

$\displaystyle t = \frac{\sqrt{V^2 + 2 g h}}{g} $

Total time taken $\displaystyle = \frac{\sqrt{V^2 + 2 g h}}{g} + \frac{V}{g} $