Question

a rock is thrown straight up with an initial velocity of 30.0 m/s.what maximum height will the rock reach before starting to fall downward? (acceleration due to gravity is 9.80 m/s2.)

Answer 1

V = a t t is the same upwards or downwards because of conservation of energy

t = V / a = 30.0 m/s / 9.80 m/s^2 = 3.06 sec it will take 3.06 sec for a falling rock to reach 30.0 m/s

H = 1/2 g t^2 = 1/2 * 9.80 m/s^2 * 3.06^2 s^2 = 45.9 m

Note: one could calculate

H = Vo t - 1/2 g t^2 where t = 30.0 m/s / 9.80 m/s = 3.06 sec and one should obtain the same result