A ball is dropped on the floor from a height of 20 m . It rebounds to a height of 5 m . Ball remains in contact with floor for 1 s . The average acceleration during contact is (acceleration due to gra

Question

A ball is dropped on the floor from a height of 20 m . It rebounds to a height of 5 m . Ball remains in contact with floor for 1 s . The average acceleration during contact is (acceleration due to gravity = 10 m / s 2 )

TestHub · Accepted Answer

Velocity of the ball just before hitting the floor is given by $v_{1} = 2 g h_{1}$ .

Substituting the given values, $v_{1} = 2 \times 10 \times 20 = 20 m/s$ (downwards).

Velocity of the ball just after rebounding is given by $v_{2} = 2 g h_{2}$ .

Substituting the values, $v_{2} = 2 \times 10 \times 5 = 10 m/s$ (upwards).

Taking the upward direction as positive, the change in velocity is $Δ v = v_{2} - (- v_{1}) = 10 - (- 20) = 30 m/s$ .

The average acceleration during contact is $a = \frac{Δ v}{Δ t}$ .

Substituting $Δ t = 1 s$ , we get $a = \frac{30}{1} = 30 m/s^{2}$ .

Answer: $30 m / s^{2}$

Physics - NLM Question with Solution | TestHub

Options:

Doubts & Discussion