Answer:
3.055 m
Step-by-step explanation:
In this solution we will use next notation:
[tex]t_1[/tex]= time elapsed since oco serves the ball until it reaches its opponent.
[tex]t_2[/tex]= 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
[tex]t_1=\frac{25m}{50m/s}=0.5s[/tex]
To compute t_2 observe that the return speed is 12.5 m/s and the distance that the ball will travel is [tex]25-(10t_1+10t_2)[/tex]. Then,
[tex]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[/tex].
Therefore,
[tex]d=10(t_1+t_2)=10(0.5+\frac{8}{9})=10(\frac{17}{18})=\frac{85}{9}m[/tex]
Finally, as Oco started 12.5m away from the net, when she meets the return she will be
[tex]12.5-\frac{85}{9}=\frac{55}{18}=3.055m[/tex]
away from the net.
