A ball is thrown vertically downwards with velocity $\sqrt{2 g h}$ from a height $h$ . After colliding with the ground it just reaches the starting point. Coefficient of restitution is :

1 / \sqrt{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 / 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $1 / \sqrt{2}$
The ball has an initial velocity as $\sqrt{2 g h}$
Thus the velocity just before it collides on the ground is given using $v^{2} = u^{2} + 2 a s$
or
$v^{2} = 2 g h + 2 g h = 4 g h$
or
$v_{i} = \sqrt{4 g h}$
Now the velocity changes to $v_{f}$ (say) after collision.
The ball just reaches the initial height h.
Thus we get
$0 = v_{f}^{2} - 2 g h$
or
$v_{f} = \sqrt{2 g h}$
Thus the coefficient of restitution is given as

$e = \frac{v_{f}}{v_{i}} = \frac{\sqrt{2 g h}}{\sqrt{4 g h}} = \frac{1}{\sqrt{2}}$

Suggest Corrections

Similar questions

5 m

10 m / s

(g = 10 m s^{- 2})

{10 m s}^{- 1}

{30 m s}^{- 1}

(T a k e g = {10 m s}^{- 2})

x - y

(4^i - 4^j) m / s

e = \frac{1}{2}

20 m

υ_{0}

50

υ_{0}

g = 10 m s^{- 2}

Join BYJU'S Learning Program