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September 1, 2014

Homework Help: Motion equation

Posted by Mac on Sunday, December 17, 2006 at 5:46am.

A stone, which weighs 0.62kg, is thrown, at 5.5ms, its kinetic energy at its maximum height above the surface is 2.3J, what is the height of the stone at this point? Any eqautions or advice gratefully recieved.

I assume that is 5.5 m/s, the initial velocity.

Ok, assuming that, then you know the initial KEnergy; 1/2 .62 (5.5)^2 joules.

At the max height, the KE you are given is KE from horizontal motion. So subtract that amount (2.3J) from the initial KE, and you have then the KE that can go into vertical motion. From that, it is easy:

KE available for vertical motion=PE at top, where PEnergy at the top is mass*g*height and you solve for height.

Excellent, thanks very much

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