Question

Two cannonballs, A and B, are fired from the ground with identical initial speeds, but with${\theta }_A$larger than${\theta }_B$.(a) Which cannonball reaches a higher elevation?(b) Which stays longer in the air?(c) Which travels farther? Explain.

Short answer

The obtained solution is that cannonball A will reach a higher elevation and remain longer in the air.

Step-by-step solution

Step 1. Projectile motion

In the case of a projectile motion, the vertical speed of the object is zero at the top of its path.

Between the two cannonballs A and B, cannonball A has a larger angle, and it will reach a higher elevation because the initial vertical velocity is higher.

Thus, cannonball A will reach a higher elevation.

Step 2. Determine which cannonball will remain longer in the air

As the initial vertical velocity is higher for cannonball A, it will take more time to decelerate to zero and then begin to fall. Also, cannonball A has a larger angle; therefore, it will remain longer in the air.

Thus, cannonball A will remain longer in the air.

Step 3. Determine which cannonball will travel farther

Between the two cannonballs A and B, the ball having the launch angle closest to $45°$will move the farthest. The range is the maximum for a launch angle having $45°$, and it reduces for the angles more or less than $45°$.

Thus, the cannonball with the launch angle closest to $45°$will travel the farthest.