A cannon, located 60.0 m from the base of a vertical 25.0 -m-tall 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

muzzle velocity = s
U = horizontal speed = s cos 43 forever = .731 s
time to cliff = T = 60/u = 60/(.731 s) = 82.1 /s
h = s sin 43 * t - 9.81/2 t^2
when t = T, h = 25
25 = s sin 43 (82.1/s) - 4.9 (82.1^2/s^2)
25 = 56 - 33028/s^2
33028 / s^2 = 31
s^2 = 1065
s = 32.6 meters/s
Now find the second time it reaches h = 25
h = s sin 43 * t - 9.81/2 t^2
25 = 32.6 sin 43 t - 4.9 t^2
4.9 t^2 - 22.2 t + 25 = 0
solve quadratic for the first and second t. The second t is when it lands back on the cliff.