Acannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0° 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?

25.2 m

Explanation:

The horizontal distance traveled s = v × t×cosθ    and t = s / (v×cos θ)

The vertical distance traveled h = v × t ×sin θ - 1/2 × g × t^2

Substituting for t,  h = s×tan θ- 1/2 × g × s^2 / (v cos θ)^2

Now, Solve for v^2   and get v^2 = g × s^2 ÷ [2 ×cos^2θ × (s×tan θ - h)]

And v^2 = 9.8×3600 / [2×0.535×(60×0.933 - 25] =1065    and v = 32.6 m/s

As a check from the first equation t = 60 / (32.6×0.731) = 2.52 sec

Horizontal distance traveled s = 32.6×cos 43°×2.52 =60 m

Height reached 2×g×h = (v×sin43°)^2    

⇒and h = 25.2 m  (using 2×g×h =v×y ^2)

Since maximum height is reached at the edge of the cliff the projectile

will not travel beyond the cliff

probably not if net force is zero