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What is the ball's impact speed on the ground below?

A cannon tilted up at a 31.0 angle fires a cannon ball at 66.0m/s from atop a 25.0m -high fortress wall.

What is the ball's impact speed on the ground below?

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The Best answer for What is the ball's impact speed on the ground below?

Using Conservation of Energy:

Ek + Eg = Ek

1/2 m vi^2 + mgh = 1/2 m vf^2

vf = sqrt (vi^2 + 2gh)

vf = sqrt (66^2 + 19.6(25))

vf = 69.6 m/s

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In the two situations, the preliminary complete ability of the ball is m*h + ½*m*v²; so the kinetic ability whilst the ball hits the floor is an identical, and the speed is an identical. to seem at it yet in a distinctive way, if the ball is thrown up with speed v, it ought to have this comparable speed interior the downward direction whilst it passes the altitude of the cliff. for this reason the tip result's comparable to if the ball replaced into thrown downward with that speed.

i think this is the correct answer for What is the ball's impact speed on the ground below?

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The initial kinetic energy is:

KEi = (1/2)m(vi)^2

The final kinetic energy is:

KEf = (1/2)m(vf)^2

The initial potential energy is:

PEi = mgh

The final potential energy is:

PEf = mg(0) = 0

Since energy is conserved:

KEf + PEf = KEi + PEi

(1/2)m(vf)^2 + 0 = (1/2)m(vi)^2 + mgh

Multiply both sides by 2 and divide both sides by m:

vf^2 = vi^2 + 2gh

Take the square root of both sides, and plug in given values:

vf = sqrt[vi^2 + 2gh] = sqrt[(66.0 m/s)^2 + 2(9.81 m/s^2)(25.0 m)] = 69.6 m/s

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