we shoot a pumpkin horizontally from a cannon at a height of 412 ft with a...

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we shoot a pumpkin horizontally from a cannon at a height of 412 ft with a...

we shoot a pumpkin horizontally from a cannon at a height of 412 ft with a velocity of 50 m/s South. whats the potential energy of the pumpkin the instant it emerges from the pumpkin cannon?

science physics

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Answer 1

Answer #1

Let 'm' be the mass of the pumpkin.

Then, potential energy of pumpkin just outside canon,

PE = mgh = m * 9.81 * (412 * 0.3048) = [1231.9 * m] J

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Answer 2

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