What is the initial velocity of a rock thrown off a 20-m high cliff?

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A rock thrown horizontally from a 20-meter high cliff strikes the ground at a 45-degree angle, prompting a discussion on calculating its initial velocity. The key equations involve the range and velocity components, with the final y-velocity derived from gravitational acceleration. The relationship between x and y components is crucial, as the x-velocity remains constant while the y-velocity is influenced by gravity. After analyzing the components, the initial velocity is determined to be 20 m/s. Understanding the motion visually rather than solely through equations aids in solving the problem effectively.
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Homework Statement


A rock is thrown horizontally from the top of a 20-m high cliff and strikes the ground at an angle of 45 degrees. With what speed was it thrown?


Homework Equations


R = V2/g*sin(2theta)
v2 = (v0*sin(theta))2 - 2gy

The Attempt at a Solution



I have 2 unknowns no matter what equation I try to use. I've tried to get it down to one unknown, but I keep getting the wrong answer..
 
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Don't think in equations, think in terms of the picture you see. You grabbed the range equation, but don't just try to plug things into it.

The rock strikes the ground at 45 degrees. That means it's final velocity has x and y components that are related to each other how? You know x velocity is constant. You can find final y velocity using that initial y velocity is zero. If you know the relation between x and y velocity, you can then find x velocity.
 
Ahhh, I think I understand now.

v2 = 0 - 2gy
v = (2*9.8*20)1/2
= 19.7

The correct answer is 20m/s. Thanks :smile:
 
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