Need help with physics hw!?

Question

A player passes a 0.6 kg basketball downcourt for a fast break. The ball leaves the player's hands with a speed of 7 m/s and slows down to 4.9 m/s at its highest point.

How high (above the release point) is the ball when it is at its maximum height?

Answer 1

the vertical speed is
vy=sqrt(49-4.9^2)

and
at apogee
t=vy/g
so
y=.5*vy^2/g

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