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?

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)

Taking the net as the origin (x = 0), , velocity will be nagative and assuming there is no friction wit air acceleration would be 0, so:

(2)

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

(3)

So:  

The opponent will return the ball at , the equation for the return of the ball will be:

(4)

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

(5)

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

(6)

As in the first case it starts from 25m, , acceleration equals 0 and velocity is negative :

(7)

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

Replacing the values:

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

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

the answer is a)

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multiply 2kg by latent heat of vap