posted by Bojian .
A tennis ball on Mars, where the acceleration due to gravity is 0.379 of a and air resistance is negligible, is hit directly upward and returns to the same level 6.20 later.
(a)How high above its original point did the ball go?
(b)How fast was it moving just after being hit?
g = .379(9.8) = 3.71m/s^2
Up time = down time = 3.1sec. each.
Vf = 0 = Vo - 3.71(3.1) making
Vo = 11.5m/s
h = 11.5(3.1) - 3.71(3.1)^2/2 = 17.82m
(b) Vo = 11.5m/s.