# The impulse of a tennis ball

1. Jul 28, 2012

### Northbysouth

1. The problem statement, all variables and given/known data
To warm up for a match, a tennis player hits the 58.0g ball vertically with her racket.
If the ball is stationary just before it is hit and goes 5.50m high, what impulse did she impart to it?

I know that the answer is 0.602 kgm/s but I don't understand why this is the answer
2. Relevant equations
P=mv

ptotal = Pinitial + Pfinal

3. The attempt at a solution
Pinitial = (0.058kg)(0 m/s)
= 0

Pfinal = (0.058kg)(vfinal

I'm not sure where to go from here.
I think that kinitial+Uinitial = kfinal+ufinal has something to do with it
But I'm not sure.

2. Jul 28, 2012

### cepheid

Staff Emeritus
Energy is conserved. If the ball reaches a height of 5.50 m, it has gained a certain amount of gravitational potential energy (you can calculate the amount). That energy had to come from somewhere. From this fact, you can deduce the speed of the ball at the end of the impact with the racket. ;)

3. Jul 29, 2012

### fredb

compute v(0) tennis ball: v(t)[/2] = v(0)[/2] - 2*g*d → 0 = v(0)[/2] - 2*9.8*5.5 = 10.38 m/s.

ƩP(initial) = ƩP(final) (just before and just after the hit). In general p=mv.

Initial m(b)*v(b) + m(r)*v(r) = 0+ m(r)*v(r)

Final m(b)*v(b) + m(r)*v(r) = 0.058*10.38=0.602 kgm/s

So initial momentum racket is 0,602 kgm/s.

4. Jul 29, 2012

### cepheid

Staff Emeritus
Fredb, we do NOT provide complete solutions to homework problems on this site. It is against site rules.