A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0m above a flat, horizontal beach. With what angle of impact does the stone land (in degrees).

Answer 1

v = Vo -9.8 t
z = h + Vo t - 4.9 t^2
here Vo = initial speed up = 0
h = 50
z = final height = 0
so
0 = 50 - 4.9 t^2
so t = 3.19 seconds to fall
v = 0 - 9.8 * 3.19 = -31.3 m/s
so the speed down is 31.3 and the horizontal speed is still 18
tangent of angle from vertical = 18/31.3