Final answer:
The minimum muzzle velocity for the shell to clear a 15.0 m tall cliff can be determined using projectile motion equations. The motion is analyzed by separating the problem into vertical and horizontal components, applying kinematic equations to each.
Step-by-step explanation:
To calculate the minimum muzzle velocity needed for a cannonball to clear a cliff, we can apply the principles of projectile motion. Given the angle of 43.0 degrees above the horizontal, and the need to clear a 15.0 m vertical cliff while the cannon is 60.0 m away, we have a classic application of the physics of two-dimensional motion.
We can break this problem down by considering the vertical and horizontal components of the projectile's motion separately. The vertical component must ensure that the projectile reaches a height of 15 m, while the horizontal component must account for the distance to the cliff.
The relevant equations for the vertical motion are those dealing with uniformly accelerated motion, such as vy = v0y + ayt and y = y0 + v0yt + (1/2)ayt2, where vy is the final vertical velocity, v0y is the initial vertical velocity, ay is the acceleration due to gravity, and t is the time in flight. The horizontal motion can be described simply by x = v0xt, where x is the horizontal displacement, and v0x is the initial horizontal velocity.
To solve for the minimum muzzle velocity, we must find the initial vertical velocity that will exactly reach a height of 15 m, as well as the corresponding horizontal velocity needed to travel 60 m before gravity drops the projectile below the top of the cliff.