A tennis ball is dropped on to the floor from a height of 9 . 8 m . It rebounds to a height 5 . 0 m . Ball comes in contact with the floor for 0 . 2 s . The average acceleration during contact is ____

Question

A tennis ball is dropped on to the floor from a height of 9 . 8 m . It rebounds to a height 5 . 0 m . Ball comes in contact with the floor for 0 . 2 s . The average acceleration during contact is ______ m s - 2 . [Given g = 10 m s - 2 ]

TestHub · Accepted Answer

The speed of ball just before collision with ground is $v_{i}^{2} = 0 + 2 g h_{i}$

$\Rightarrow v_{i} = \sqrt{2 g h_{i}}$

$= \sqrt{2 \times 10 \times 9.8}$

$= 14 m s^{- 1} (Downward)$

The speed of ball just after collision is $0 = v_{f}^{2} - 2 g h_{f}$

$\Rightarrow v_{f} = \sqrt{2 g h_{f}}$

$= \sqrt{2 \times 10 \times 5}$

$= 10 m s^{- 1} (Upward)$

Average acceleration of ball is $|{\vec{a}}_{avg}| = |\frac{Δ \vec{v}}{Δ t}| = \frac{v_{f} - (- v_{i})}{t} = \frac{10 + 14}{0.2} = \frac{24}{0.2} = 120 m s^{- 2}$ .

Physics - Kinematics Question with Solution | TestHub

Doubts & Discussion