First slide

Average Acceleration

Question

A ball is dropped on to the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with floor for 0 .01 second, what is the average acceleration during contact ?

Moderate

A

700 m/s²
B

1400 m/s²
C

2100 m/s²
D

2800 m/s²

Solution

According to above question,
$\begin{array}{l} v_{1} = \sqrt{(2 gh)} = \sqrt{2 \times 10 \times 10} = \sqrt{200} \\ v_{2} = - \sqrt{(2 gh)} = - \sqrt{2 \times 10 \times 2 \cdot 5} = - \sqrt{50} \end{array}$
So, $Δv = v_{1} - v_{2} = \sqrt{(200)} + \sqrt{(50)} = 3 \sqrt{50} \approx 21$
Acceleration $= \frac{ΔV}{Δt} = \frac{21}{0 \cdot 01} = 2100 m / s^{2}$ .

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App