Physics
posted by Elisa .
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 15.0 m/s. The cliff is h = 66.0 m above a flat horizontal beach.
How long after being released does the stone strike the beach below the cliff?
I got 3.67 s.
With what speed and angle of impact does the stone land?
I got 38.97 m/s for the speed, but I cannot figure out the angle. It says that the answer is in degrees "below the horizontal."
I calculated x to be 55.23m and I tried to calculate y, but I keep getting the wrong angle. After getting y I though I was supposed to do arctan(opp/adj) to get the angle "below the horizontal."
Please help! Thanks!

T = 3.67 s is correct for the time to impact. When thowing horizontally, the speed does not affect the time to hot the ground.
Using conservation of energy for the speed Vf at impact,
(1/2) Vf^2 = Vo^2/2 + g H
(1/2) Vf^2 = (15^2)/2 + 9.8*66
= 759.3 m^2/s^2
Vf = 38.97 m/s
The tangent of the angle at impact, measured below the horizontal, is Vy/Vx at impact
theta = arctan (g*T)/15 m/s
= arctan 35.97/15 = 67.4 degrees 
Thanks!

It is also arcsin g*T/38.97, where 38.97 is the speed. This gives you the same angle.
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