Question

Coco 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

Distance from the net when she meets the return: 14,3 m

Step-by-step explanation:

First we need to know how much time it takes the ball to reach the net, the kinematycs general equation for position:

(1)
`x = x_(0)  + V_(s)  t + (1)/(2)  a t^(2)`

Taking the net as the origin (x = 0),
`x_(0) = 25m`, velocity will be nagative
`V_(s) = - 50m/s` and assuming there is no friction wit air acceleration would be 0, so:

(2)
`x = x_(0)  + V_(s)  t`

we want to know the time when it reaches the net so when x=0, replacing de values:

(3)
`0 = 25m  - 50 m/s t_(net)`

So:
`t_(net)  = 0,5 s`

The opponent will return the ball at
`V_(ret) = 25m/s`, the equation for the return of the ball will be:

(4)
`y = y_(0)  + V_(ret)  t + (1)/(2)  a t^(2)`

Note that here it start from the origin,
`y_(0) = 0`, as in the other case acceleration equals 0, and here we have to consider that the time starts when the ball reaches the net (
`t_(net)`) so the time for this equiation will be
`t - t_(net)`, this is only valid for
`t >= t_(net)`:

(5)
`y = V_(ret)  (t - t_(net))`

Coco starts running as soon as he serves so his equiation for position will be:

(6)
`z = z_(0)  + V_(c)  t + (1)/(2)  a t^(2)`

As in the first case it starts from 25m,
`z_(0) = 25m`, acceleration equals 0 and velocity is negative
`V_(c) = - 10m/s`:

(7)
`z = z_(0) + V_(c)  t`

To get the time when they meet we have that
`z = y`, so from equiations (5) and (7):

`V_(ret)  (t - t_(net)) =  z_(0) + V_(c)  t`

`t = (z_(0) + V_(ret)* t_(net))/(V_(ret)-V_(c))`

Replacing the values:

`t = 1,071 s`

Replacing t in either (5) or (7):

`z = y = 14,3 m`

This is the distance to the net when she meets the return