Question

oco serves a tennis ball at vs = 50 m/s and charges the net at vc = 10 m/s. The opponent, x = 25 m away on the other side of the court, returns the ball with a speed half that of the serve. How close does Coco get to the net (x/2 away) before she meets the return?

Answer 1

3.055 m

Explanation:

In this solution we will use next notation:

`t_1`= time elapsed since oco serves the ball until it reaches its opponent.

`t_2`= time elapsed since the opponent returns the ball until it reaches oco.

d= Total distance traveled by Oco since serving the ball until meeting the return.

We know that oco serves at vs = 50 m/s and her opponent is x=25 m away. Then, t_1 is given by

`t_1=(25m)/(50m/s)=0.5s`

To compute t_2 observe that the return speed is 12.5 m/s and the distance that the ball will travel is
`25-(10t_1+10t_2)`. Then,

`t_2=(25-10t_1-10t_2)/(12.5)=(20-10t_2)/(12.5)\implies t_2=(20)/(22.5)=(8)/(9)s`.

Therefore,

`d=10(t_1+t_2)=10(0.5+(8)/(9))=10((17)/(18))=(85)/(9)m`

Finally, as Oco started 12.5m away from the net, when she meets the return she will be

`12.5-(85)/(9)=(55)/(18)=3.055m`

away from the net.