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A cannon tilted up at a 34.0 ∘ angle fires a cannon ball at 84.0 m/s from atop a 27.0 m -high fortress wall.?
What is the ball's impact speed on the ground below?
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- ?Lv 73 years agoFavorite Answer
Since we need only the magnitude of the velocity (and not the angle), this is most easily done using conservation of energy:
KE + PE at top of wall = KE at ground
½mu² + mgh = ½mv²
u² + 2gh = v²
(84.0m/s)² + 2 * 9.8m/s² * 27.0m = v²
v = 87.1 m/s
Hope this helps!
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