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

Answer:

3.055 m

Step-by-step 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

$t_1=\frac{25m}{50m/s}=0.5s$

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,

$t_2=\frac{25-10t_1-10t_2}{12.5}=\frac{20-10t_2}{12.5}\implies t_2=\frac{20}{22.5}=\frac{8}{9}s$.

Therefore,

$d=10(t_1+t_2)=10(0.5+\frac{8}{9})=10(\frac{17}{18})=\frac{85}{9}m$

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

$12.5-\frac{85}{9}=\frac{55}{18}=3.055m$

away from the net.