Physics
posted by Ana on .
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 19.0 m/s. The cliff is h = 50.0 m above a flat horizontal beach.
With what speed and angle of impact does the stone land?

find the vertical speed at impact:
vf^2=2*g*h
You know the vertical velocity, and the horizontal velocity.
From that, you get the angle, and the magnitude. 
I was able to find out the angle which is 57.83degree that is correct.
For the speed, i used sqrt(2*9.81*50)=31.32, is this correct because the hw site doesn't accept it. What am I missing?
Thanks. 
The sqrt(2*9.81*50)= 31.32 (m/s) that you calculated is the vertical velocity compnent. There is also a horizontal component (19.0 m/s)
You need to combine them with the Pythagorean equatiuon to get the speed.