How Is Impulse Calculated in Tennis Ball Physics?

[Northbysouth]

Homework Statement

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

Homework Equations

P=mv

p_total = P_initial + P_final

The Attempt at a Solution

P_initial = (0.058kg)(0 m/s)
= 0

P_final = (0.058kg)(v_final

I'm not sure where to go from here.
I think that k_initial+U_initial = k_final+u_final has something to do with it
But I'm not sure.

 

[cepheid]

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. ;)

 

[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.

 

[cepheid]

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