Physicsss!!...help!?

Question

A cannon, located 60.0 m from the base of a vertical 25.0 m tal cliff, shoots a 15 kg shell at 43.0 degrees above the horizontal toward the cliff.

a) What must the minimum muzzle velocity be for the shell to clear the top of the cliff?

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?

Answer 1

Set up the coordinate system with unit m, such that the edge of the cliff is at (0, 0). Accordingly, the cannon is at (-60.0, -25.0).
(a) Let the minimum muzzle velocity be V (m/s). The horizontal and vertical componenets of this muzzle velocity V are: V*cos(43.0°) and V*sin(43.0°), respectively.
Since the horizontal would not be influenced by gravity and thus is a constant, the time for the 15 kg shell to clear the edge of the cliff is: t = 60.0/ (V*cos(43.0°))
With this t known, and with the minimum muzzle velocity, we have:
V*sin(43.0°)*t - 0.5g*t^2 = 25.0
or: 60.0*tan(43.0°) - 0.5*g*60.0^2/{V^2*(cos(43.0°))^2} = 25.0
V^2 = 0.5*g*60.0^2/ {(60.0*tan(43.0°) - 25.0)*(cos(43.0°))^2}
V = (60.0/cos(43.0°))*sqrt{0.5*g /(60.0*tan(43.0°) - 25.0)}
= 32.66 (m/s)
(b) Put this V back into the formula for t, we get:
t = 60.0/ (V*cos(43.0°)) = 2.512 (s)
Also, set T to be the time for the shell to reach the maximum height, we have: T = V*sin(43.0°)/g = 2.27 (s)
This indicates that the shell reaches its maximum height before reach the edge of the cliff. Therefore the answer to (b) is zero (0.00)m.