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} $