Minimum velocity question

  • Thread starter ledhead86
  • Start date
  • #1
ledhead86
59
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! 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
 

Answers and Replies

  • #2
Poncho
29
0
Don't panic.
 
  • #3
ledhead86
59
0
Poncho said:
Don't panic.
Thanks for the help. I completely understand now.
 
  • #4
Poncho
29
0
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.
 

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