Question

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0ms, then what is the average acceleration during that contact?

• A. - 1.26 × 10³ m/s²
• B. + 10 m/s²
• C. - 10 m/s²
• D. + 1.26 × 10³ m/s²

Answer 1

The correct option is D

+ 1.26 × 10³ m/s²

u = 0 m/s since ball is dropped

s = -4 m since displacement in negative y direction

a = 10 $m/s^2$

$v^2=2(-10)(-4)=80$

$\Rightarrow v=\pm \sqrt{80}$

$\because$velocity in negative y so v = $-\sqrt{80}$ m/s

Now the ball rebounds with some velocity

Since the collision is inelastic the ball won't rebound with some velocity.

Don't worry you will come to know about elastic and inelastic later, it's easy.

We know the ball rises up to a height of 2 m. So,

s = + 2m

v = 0 m/s as finally it stops

a = -10 $m/s^2$

So,

$v^2-u^2$ = $2a\lambda$

$-u^2=2(-10)2$

$=\pm \sqrt{40}$

Since u is in positive y direction so u should be positive or $\sqrt{40}$

so just before coming in contact ball's velocity was $-\sqrt{80}$ and just after it became $+\sqrt{40}$ m/s

Time duration is given as 0.012

So average acceleration = $\frac{\text{change in velocity}}{\text{time taken}}$

$=\frac{\sqrt{40}-(-\sqrt{80})}{0.012}$

$=1.27\times {10}^{3}m/s^2$

The acceleration should be positive as velocity is being changed from negative to positive.