A cannonball is fired into the air with a speed of 42 m/s from a height of 28 m, with what speed does the cannonball hit the ground?

To find the speed with which the cannonball hits the ground, we can use the equations of motion.

Step 1: Determine the initial velocity (u), the final velocity (v), the acceleration (a), and the displacement (s).

In this case, the initial velocity is given as 42 m/s, the acceleration is due to gravity which is about 9.8 m/s² (assuming no air resistance), and the displacement is equal to the height from which the cannonball is fired, which is 28 m. We need to find the final velocity.

Step 2: Use the equation of motion relating u, v, a, and s:

v² = u² + 2as

Plugging in the values:

v² = (42 m/s)² + 2 * (-9.8 m/s²) * (28 m)

Step 3: Solve for v by taking the square root of both sides of the equation.

v = √[(42 m/s)² + 2 * (-9.8 m/s²) * (28 m)]

Calculating this value will give us the speed at which the cannonball hits the ground.