A man stands on the roof of a building of height 15.8m and throws a rock with a velocity of magnitude 25.2m/s at an angle of 32.7∘ above the horizontal. You can ignore air resistance.

Question

Part A
Calculate the maximum height above the roof reached by the rock.
y =
9.46mCorrect
Part B
Calculate the magnitude of the velocity of the rock just before it strikes the ground.

Answer 1

h = 15.8 + (25.2 sin 32.7°)t - 4.9t^2
= 15.8 + 13.61t - 4.9t^2

I get max height = 25.25 at t=1.388, which is as you say, 9.46m above the roof.

h=0 when t=3.66
v = 13.61 - 9.8t = -22.26 when h=0