Answer:
Distance from the net when she meets the return: 14,3 m
Explanation:
First we need to know how much time it takes the ball to reach the net, the kinematycs general equation for position:
(1) [tex]x = x_{0} + V_{s} t + \frac{1}{2} a t^{2}[/tex]
Taking the net as the origin (x = 0), [tex]x_{0} = 25m[/tex], velocity will be nagative [tex]V_{s} = - 50m/s[/tex] and assuming there is no friction wit air acceleration would be 0, so:
(2) [tex]x = x_{0} + V_{s} t [/tex]
we want to know the time when it reaches the net so when x=0, replacing de values:
(3) [tex]0 = 25m - 50 m/s t_{net} [/tex]
So: [tex]t_{net} = 0,5 s [/tex]
The opponent will return the ball at [tex]V_{ret} = 25m/s[/tex], the equation for the return of the ball will be:
(4) [tex]y = y_{0} + V_{ret} t + \frac{1}{2} a t^{2}[/tex]
Note that here it start from the origin, [tex]y_{0} = 0[/tex], as in the other case acceleration equals 0, and here we have to consider that the time starts when the ball reaches the net ([tex]t_{net} [/tex]) so the time for this equiation will be [tex]t - t_{net} [/tex], this is only valid for [tex]t >= t_{net} [/tex]:
(5) [tex]y = V_{ret} (t - t_{net}) [/tex]
Coco starts running as soon as he serves so his equiation for position will be:
(6) [tex]z = z_{0} + V_{c} t + \frac{1}{2} a t^{2}[/tex]
As in the first case it starts from 25m, [tex]z_{0} = 25m[/tex], acceleration equals 0 and velocity is negative [tex]V_{c} = - 10m/s[/tex]:
(7) [tex]z = z_{0} + V_{c} t[/tex]
To get the time when they meet we have that [tex]z = y[/tex], so from equiations (5) and (7):
[tex]V_{ret} (t - t_{net}) = z_{0} + V_{c} t[/tex]
[tex]t = \frac{z_{0} + V_{ret}* t_{net}}{V_{ret}-V_{c}}[/tex]
Replacing the values:
[tex]t = 1,071 s[/tex]
Replacing t in either (5) or (7):
[tex]z = y = 14,3 m[/tex]
This is the distance to the net when she meets the return
