Chapter 2, Problem 12Q

A baseball player hits a ball straight up into the air. It leaves the bat with a speed of 120 km/h. In the absence of air resistance, how fast would the ball be traveling when it is caught at the same height above the ground as it left the bat? Explain.

To determine

The speed of the ball in absence of air resistance.

Answer to Problem 12Q

Solution:

  $\text{120 km/h}$

Explanation of Solution

Given:

Initial speed of ball = $\text{120 km/h}$

Formula Used:

The third equation of motion is:

  $v^2=u^2+2as$

Where, v is the final velocity, u is the initial velocity, s is the displacement and a is the acceleration.

Calculation:

As there is no air resistance, it means that the ball will experience only the gravitational force and will undergo a free fall. Acceleration remains same for the freely falling objects whereas, in presence of air, the acceleration varies.

Considering downwards as negative direction and upwards as positive.

Acceleration due to gravity = $-g$

As the ball is caught at the same height, the displacement is $s=0$ .

  $\begin{array}{l}{v}^2=u^2+2as\\ v^2={\left(120\right)}^2+0\\ \left|v\right|=120\text{km}\end{array}$

Thus, the magnitude of the final velocity i.e. the final speed of the ball is same as its initial speed.

Conclusion:

In the absence of air resistance, the falling objects has the same initial and final speed when caught at the same height.