Posted by **Ana** on Saturday, February 13, 2010 at 5:56pm.

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?

- Physics -
**bobpursley**, Saturday, February 13, 2010 at 6:18pm
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.

- Physics -
**Ana**, Saturday, February 13, 2010 at 6:26pm
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.

- Physics -
**drwls**, Saturday, February 13, 2010 at 7:20pm
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.

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