Question

The fastest server in women's tennis is Sabine Lisicki, who recorded a serve of 131 mi/h (211 km/h) in 2014. Suppose that the acceleration of the ball was constant during the contact with the racket. Part A If her racket pushed on the ball for a distance of 0.15 m, what was the acceleration of the ball during her serve?

Answer 1

Answer:

The acceleration of the ball during her serve is 82854.03 m/s².

Explanation:

Given that,

Speed = 211 km/h

Distance = 0.15 m

We need to calculate the time

Using formula of speed

$v=\dfrac{d}{t}$

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{0.15}{211\times\dfrac{5}{18}}$

$t=0.00255\ sec$

We need to calculate the acceleration

Using formula of acceleration

$v= u+at$

$a=\dfrac{211\dfrac{5}{18}-0}{0.00255}$

$a=82854.03\ m/s^2$

Hence, The acceleration of the ball during her serve is 82854.03 m/s².