Chapter 3.3, Problem 43PP

To determine

The velocity of the ball when it hits the sidewalk.

Answer to Problem 43PP

  $8.281\text{m/s}$

Explanation of Solution

Given:

A ball is dropped from a window $3.5\text{ m}$ above the sidewalk.

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:

Since, the ball is dropped from the window, thus its initial velocity will be 0, that is,

  $v_{initial}=0\text{m/s}$

Acceleration acting on it 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 period of the motion will be:

  $\begin{array}{c}{d}_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ 3.5=0+0\left(\Delta t\right)+\frac{1}{2}\left(9.8\right){\left(\Delta t\right)}^2\\ \Delta t=0.845\text{s}\end{array}$

Then, the final velocity of the ball will be,

  $\begin{array}{c}{v}_{final}=v_{initial}+a\Delta t\\ =0+\left(9.8\right)\left(0.845\right)\\ =8.281\text{m/s}\end{array}$

Conclusion:

Thus, the final velocity of the ball when it hits the sidewalk is $8.281\text{m/s}$ .