Question

A projectile is fired from the top of a cliff at 14.2 m/s at an angle of 36.9 degrees below the horizontal, and strikes water at the same level as the base of the cliff 3.51 seconds later. (Assume g=10.0 m/s2.)
How high was the top of the cliff above the water?

Answer 1

all we care about is the downward velocity. Initially that is

14.2 cos -36.9° = -11.36 m/s

so, starting from height H, the height after t seconds is

-11.36t - 4.9 t^2
at t=3.51,
h = -100.24 m

So, the cliff is 100.24m high