Chapter 3.3, Problem 44PP

a.

To determine

The height to which the ball rises.

Answer to Problem 44PP

  $25.83\text{ m}$

Explanation of Solution

Given:

Initial speed of the ball $u=22.5\text{ m/s}$

Formula used:

According to third equation of motion.

  $v^2=u^2+2ah$

Here, v is final velocity, u is initial velocity, a is acceleration ands is displacement.

Calculation:

  $v^2=u^2+2gh$

Here, $a=g$ is acceleration due to gravity and $s=h$ is height.

When the ball is at maximum height, speed of the ball will be 0.

Therefore, final velocity $v=0\text{ m/s}$

As the ball is thrown upward, so acceleration due to gravity is $g=-9.8{\text{ m/s}}^2$ .

Now, Substitute the valuesand solve:

  $\begin{array}{l}{0}^2={22.5}^2+2\times \left(-9.8\right)\times h\\ h=\frac{{22.5}^2}{2\times 9.8}\\ =25.83\text{ m}\end{array}$

Conclusion:

Therefore, height through which ball rises is 25.83 m.

b.

To determine

The time for which the ball will be in air.

Answer to Problem 44PP

4.58 s

Explanation of Solution

Given:

Initial speed of the ball $u=22.5\text{ m/s}$

Formula used:

  $v=u+at$

Here, v is final velocity, u is initial velocity, a is acceleration and t is time.

Calculation:

When the ball is at maximum height, speed of the ball will be 0 .

Therefore, final velocity $v=0$

As the ball is thrown upward, so acceleration due to gravity $g=-9.8{\text{ m/s}}^2$

Now, putting the value of u, v, g and t.

  $\begin{array}{c}0=22.5+\left(-9.8\right)\times t\\ t=\frac{22.5}{9.8}\\ =2.29\text{ s}\end{array}$

So, the ball will be in air for $2\times 2.29\text{s = 4}\text{.58}\text{s}$

Therefore, ball will be in air for 4.58 s.