A 0.300 kg ball is thrown vertically upward with an initial speed of 17.0 m/s. part a what is the maximum height reached by the ball?

1/2 mv^2 = mgh, all of the initial kinetic energy is converted to potential energy

So (since you can see how 'm' divides out):

(0.5) v^2 = gh, g = acceleration of gravity

(0.5) (20.0) ^2 = (9.8) h

200 = 9.8h

h=20.4 m