stone throwing speed and angle?

Question

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 17.0 m/s. The cliff is h = 61.0 m above a flat horizontal beach.



How long after being released does the stone strike the beach below the cliff?
_____s
With what speed and angle of impact does the stone land?
_____m/s
_____° below the horizontal

Answer 1

Height, Sv = 61 m
Vertical speed, Uv = 0 m/s
Horizontal speed, Uh = 17 m/s

Sv = Uv t + 1/2 g t^2
61 = 0 + 1/2 (9.81) t^2
61 = 4.905 t^2
t^2 = 12.436
time for stome strike beach, t = 3.526 s

Vh = Uh = 17 m/s
Vv = Uv + g t
Vv = 0 + 9.81(3.526)
Vv = 34.6 m/s

Horizontal Speed = 17m/s
Vertical Speed = 34.6m/s

Average Speed when stone strike land
= sqrt(17^2 + 34.6^2) = 38.55 m/s

Angle below horizonatl when stone strike land
= tan^-1 (34.6/17)
= tan^-1 2
= 63.5 degree

Answer 2

h = 0.5gt^2
t = √(2h/g) = √(2*17/9.8) = 1.86 s

v = √[17^2 + (gt)^2] = 24.93 m/s

angle = arctan(gt/17) = 47.0 degree below the horizontal.