Chapter 3, Problem 74A

To determine

The velocity of the ball when it hits the sidewalk.

Answer to Problem 74A

  $7.27\text{m/s}$

Explanation of Solution

Given:

A ball is thrown from a window $2.5\text{ m}$ above the sidewalk with an initial velocity of $2.0\text{m/s}$ .

Formula used:

Final velocity with constant acceleration can be obtained by:

  $v_{final}=v_{initial}+a\Delta t$

Where, $\Delta t$ is the time taken, a is acceleration and v represents velocity.

The distance traveled by an object with a constant acceleration can be obtained by:

  $d_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a\Delta t^2$

Here, d represents displacement, v represents velocity, t represents time and a is the acceleration.

Calculation:

Acceleration acting on the ball will be due to gravity only. Thus,

  $a=9.8{\text{m/s}}^{\text{2}}$

Now, using the above mentioned formula for the distance covered, the time elapsed of the motion will be

  $\begin{array}{l}{d}_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ 2.5=0+2\left(\Delta t\right)+\frac{1}{2}\left(9.8\right){\left(\Delta t\right)}^2\\ 4.9{\left(\Delta t\right)}^2+2\left(\Delta t\right)-2.5=0\\ \Delta t=\frac{-2\pm \sqrt{2^2-4\left(-2.5\right)\left(4.9\right)}}{2\times 4.9}\\ \Delta t=0.538\text{s}\end{array}$

Then, the final velocity of the ball will be:

  $\begin{array}{c}{v}_{final}=v_{initial}+a\Delta t\\ =2.0+\left(9.8\right)\left(0.538\right)\\ =7.27\text{m/s}\end{array}$

Conclusion:

Thus, the final velocity of the ball when it hits the sidewalk is $7.27\text{m/s}$ .