A ball is thrown vertically downwards from a height of 20 m with an initial velocity v0 . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v0 is: (Take g=10 m s- 2 )

Question

A ball is thrown vertically downwards from a height of 20 m with an initial velocity v0 . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v0 is: (Take g=10 m s- 2 ) (A) 20 m s- 1 (B) 28 m s- 1 (C) 10 m s- 1 (D) 14 m s- 1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-5hn65jovtby9.png" /> kinetic energy of the ball on reaching ground K E1 = m g h + (1/2) m v02 after collision (K E1/2) = m g h (m g h + (1/2) m v02/2) = m g h v0 = √2 g h = 20 textm texts- 1

Q. A ball is thrown vertically downwards from a height of $20 m$ with an initial velocity $v_{0}$ . It collides with the ground, loses $50$ percent of its energy in collision and rebounds to the same height. The initial velocity $v_{0}$ is: (Take $g = 10 m s^{- 2}$ )

$20 m s^{- 1}$

$28 m s^{- 1}$

$10 m s^{- 1}$

$14 m s^{- 1}$

kinetic energy of the ball on reaching ground
$K E_{1} = m g h + \frac{1}{2} m v_{0}^{2}$
after collision $\frac{K E _{1}}{2} = m g h$
$\frac{m g h + \frac{1}{2} m v _{0}^{2}}{2} = m g h$
$v_{0} = 2 g h = 20 m s^{- 1}$

Q. A ball is thrown vertically downwards from a height of 20m with an initial velocity v0​ . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v0​ is: (Take g=10ms−2 )

20ms−1

28ms−1

10ms−1

14ms−1