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# Projectile motion of shell shot from a cannon

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A cannon, located 60.0m from the base of a vertical 25.0m tall cliff, shoots a 15kg shell at 43.0 digress above the horizontal toward the cliff.

1). What must the minimum muzzle velocity be for the shell to clear the top of the cliff?

2) The ground at the top of the cliff is level, with a constant elevation of 25.0m above the cannon. Under the conditions of question 1, how far does the shell land past the edge of the cliff?

Please explain your answers (I know you can do it, I need to see how to do it)

##### Solution Summary

The solution is a detailed explanation for the projectile motion of the shell shot from a cannon, which is located at the base of a cliff. By decomposing the motion of two orthogonal components, it shows the calculation of finding the maximum vertical height the shell traveled and its horizontal displacement when landing on the cliff.

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