- #1

ledhead86

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**A cannon, located 60.0 m from the base of a vertical 25.0m-tall cliff, shoots a 15-kg shell at 43 degrees above the horizontal toward the cliff.**

I have determined:

v=47.75 m/s

v_0x= 34.92 m/s

v_oy=32.56 m/s

a_x=0

a_y=-9.8 m/s^2

x_o=0

y_o=0

x=60m

y=25m

alpha= 43 degrees

part a: Therefore, the minimum muzzle velocity for the shell to clear the top of the cliff is 32.56 m/s.

Part b: The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?

How do I find part b