Question

A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally from an elevation $4.0\text{m}$above the ground. The ball lands $25\text{m}$away. What is his pitching speed?

Short answer

The pitching speed of the person is$27.67\text{m}/\text{s}$

Step-by-step solution

Step 1. Given information

The ball is thrown from an elevated$4.0\text{m}$above the ground, the ball lands in $25\text{m}$away.

Step 2. Explanation

The period of flight of the ball is

$y=u_{y}t-\frac{1}{2}a{t}^2$

y is the vertical height from the ground, is the vertical component of the velocity of the ball, and a is the acceleration due to gravity.

$\begin{array}{l}(-4.0\text{m})=(0\text{m}/\text{s})t-\frac{1}{2}(9.8\text{m}/\text{s})t^2\\ t=\sqrt{\frac{2(-4.0\text{m})}{(9.8\text{m}/\text{s})}}\\ t=0.9035\text{s}\end{array}$

Thus, the period of flight of the ball is $0.9035\text{s}$.

The horizontal distance travelled by the ball is$x=u_{x}t$

Here, x is the horizontal distance travelled by the ball, is the initial speed of the ball in horizontal direction, and t is the duration of flight.

$\begin{array}{l}(25\text{m})=u_x(0.9035\text{s})\\ u_x=27.67\text{m}/\text{s}\end{array}$

Therefore, the pitching speed of the person is$27.67\text{m}/\text{s}$