Will the Cannonball Clear the Cliff at 25 Meters High?

[ledhead86]

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

 

[DaMastaofFisix]

Tell us exactly what the question is looking for in both parts, not what you summarized it to be. Second, describe the diagram and the setting a little better so that it is easier to help out with the visuals. As for the second part, all your have to do is plug in you calculated numbers in the projectile equations and solve for the time. once you have the time, you can determine how far over the shot went. If you're still troubled, show what you have done and WHY you are stuck, not just that "you are".

 

[quicknote]

?

To find the distance the shell lands past the edge of the cliff, we can use the equations of projectile motion:

x = x_o + v_0x * t
y = y_o + v_0y * t + (1/2)*a_y*t^2

First, we need to calculate the time it takes for the shell to reach the top of the cliff. We can use the equation for vertical displacement to do this:

y = y_o + v_0y * t + (1/2)*a_y*t^2
25m = 0 + 32.56 m/s * t + (1/2)*(-9.8 m/s^2)*t^2
Solving for t, we get t = 2.64 seconds.

Now, using this value of t, we can plug it into the equation for horizontal displacement to find the distance the shell travels past the edge of the cliff:

x = x_o + v_0x * t
x = 0 + 34.92 m/s * 2.64 s
x = 92.12 m

Therefore, the shell will land 92.12 m past the edge of the cliff.