# Stone off a cliff problem

## Homework Statement

A stone is thrown off a cliff upward at a rate of 10 m/s. It then falls from the height of the throw down into a lake 20M below. How long does it take for the stone to fall? What is its impact velocity? What is the height of the cliff?

v=vi+at
x=xi+vi+1/2at^2

## The Attempt at a Solution

x=xi+vi+1/2at^2
20=0+0+1/2(9.8 x t^2)
40=9.8t^2
t^2=4
t=2.02s

v=vi+at
v=0+(9.8)(2.02)
v=19.8 m/s

x=final distance
xi=initial distance
vi=initial velocity
v=final velocity
a=acceleration of gravity (9.8)
t=time

Can someone help me to find the height of the cliff? I got the time and impact velocity but am unsure of how to get the heigh of the cliff. Thanks! :)

## Answers and Replies

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Well the cliff would be 20m relative to the lake...Shouldn't really matter relative to anything else should it? If the cliff is 100m and the lake 80m high or if the cliff is 0m high and the lake -20m you still get the same answer.

Also remember that you want to find the time when you are 20m below the initial height. So if you take the height of the cliff to be 0m you want position -20m. Lastly remember that the acceleration of gravity is downwards and you have initial velocity. My equation of motion and elapsed time differ from yours.

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Something I didn't catch before in your equation of motion, you need half the acceleration. Not half of the product $$gt^{2}$$.