(a)
First ball
Voy = 1200sin38 = 738.8 m/s
When maximum height is reached Vy = 0
Assume free fall acceleration of gravity
a = -9.81 m/s^2
2a ( y-yo) + Voy^2 = Vy^2
Yo = 0
2ay + Voy^2 = 0
Y = -Voy^2/-2a
Y = - ( 738.8)^2/-2*9.81 = 27819.8 m
Second ball
2a(y-yo) + Voy^2 = Vy^2
Voy = 1200
2ay + Voy^2 = 0
y = -Voy^2/2a = -1200^2/(2*9.81)
y = 73394 m
(b)
first ball
E = mgh = 19 * 9.81 *27819 = 5,185,341 J
second ball
E = mgh = 19 * 9.81 * 73394.5 = 13,680,000 J