How Far Does the Shell Land Beyond the Cliff Edge?

[ledhead86]

! Please Help !

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: What must the minimum muzzle velocity be for the shell to clear the top of the cliff? answer=32.6 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

 

[Poncho]

Don't panic.

 

[ledhead86]

Don't panic.

Thanks for the help. I completely understand now.

 

[Poncho]

Figure out the velocity just as the shell clears the cliff. Then treat it as a new problem as if the edge of the cliff were your starting point. From there it's a simple 2d projectile motion problem on flat ground.