Question

In a softball game, a batter hits the ball at the velocity 32 m/s and at a 50 degree angle. What is the maximum range of the ball?

Answer 1

You can split the velocity into two parts:
the horizontal part
\[u_H = 32 \text{ ms}^{-1} \times \cos 50^\circ\]
and the vertical part
\[u_V = 32 \text{ ms}^{-1} \times \sin 50^\circ\]

First thing you have to do is calculate how long the ball flies for. This is done by looking purely at the vertical part of the motion. You have an initial velocity and a negative acceleration (g = 9.8 m/s). In vertical terms you want the displacement to be zero (i.e. the ball is back on the ground). So the equation you need to solve for time t is:

\[ u_V t - \frac{1}{2} g t^2 = 0\]

One of the answers will be t=0. Ignore that one.

The maximum range is the time of flight multiplied by the horizontal velocity.

\[u_H \times t\]