A cannon elevated at 40 degrees is fired at a wall 300m away on level ground. The initial speed of the cannonball is 89m/s How long does it take for the ball to hit the wall? At what height does the ball hit the Wall?
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Answer:
to hit wall, t=d/v t=300/89cos40 t=4.4 the height needs the position function y=-.5gt^2+Vot+So So I assume is zero, it does not specify muzzle height y=-.5gt^2+vt =.5(9.8)(4.4)+89(4.4) y=370 m hits w velocity of -gt9+v v=45.588 m/s it is just about half the total distance this shot can make
tf1234 at Answerbag.com Visit the source
Other answers
89sin40 = Vy 89cos40 = Vx t = d/v t = 300/Vx That's how long it takes to hit the wall. Use that time to solve for d = Vyt.
LeopardGecko - ACS
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