Chapter 4, Problem 55A

Hanna tosses a ball straight up with enough speed to remain in the air for several seconds.

a. What is the velocity of the ball when it reaches its highest point?

b. What is its velocity 1 s before it reaches its highest point?

c. What is the change in its velocity during this 1-s interval?

d. What is its velocity 1 s after it reaches its highest point?

e. What is the change in velocity during this 1-s interval?

f. What is the change in velocity during the 2-s interval? (Caution: velocity, not speed!)

g. What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?

To determine

Velocity of the ball when reaches at highest point.

Answer to Problem 55A

At the highest point, velocity of the ball, $v=0$ .

Explanation of Solution

Given:

A ball is thrown upward in the air with enough speed so that it acn remain in the air for several seconds.

When an object is thrown vertically upwards, it continues its motion for a while but then starts to come down.

The object changes its direction from upward to downward at the highest point. At that point, the velocity becomes zero.

Conclusion:

Hence, the velocity at the highest point is zero.

To determine

Velocity, $v_1$ before 1 sec of reaching its highest point.

Answer to Problem 55A

The velocity before 1 sec, $v_1=10\text{ m/s}$ .

Explanation of Solution

Given:

Final velocity, $v=0$

Acceleration due to gravity, $g=9.8{\text{m/s}}^2$

Initial velocity, $u$

Formula used:

The equation of motion is given by:

  $v=u-gt$

Calculation:

Calculate the time at the highest point,

  $\begin{array}{l}0=u-gt\\ t=\frac{u}{g}\end{array}$

So, before one second the time will be: $t_1=\frac{u}{g}-1$

Before 1 sec of reaching the highest point, the velocity will be: $v_1=u-g{t}_1$

Substituting all the values,

  $\begin{array}{l}{v}_1=u-g\left(\frac{u}{g}-1\right)\\ v_1=9.8\text{ m/s}\end{array}$

Conclusion:

The velocity before 1 sec, $v_1=9.8\text{ m/s}$ .

To determine

Change in velocity during 1 sec interval.

Answer to Problem 55A

Change in the velocity during one second interval is $\text{9}\text{.8 m/s}$ .

Explanation of Solution

Given:

Final velocity, $v=0$

Velocity before 1 sec, $v_1=9.8\text{ m/s}$

Calculation:

Change in the velocity during this time interval will be:

  $\begin{array}{c}\Delta v=v_1-v\\ \Delta v=9.8-0\\ =9.8\text{ m/s}\end{array}$

Conclusion:

Change in the velocity during one second interval is $\text{9}\text{.8 m/s}$

To determine

Velocity after 1 sec it reaches the highest point.

Answer to Problem 55A

The velocity after one second, $v_2=10\text{ m/s}$ .

Explanation of Solution

Given:

Final velocity, $v=0$

Acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

Velocity before 1 sec, $v_1=9.8\text{ m/s}$

Formula used:

The equation of motion is given by:

  $v=u-gt$

Calculation:

After 1 second of reaching the highest point, the ball will be at the same position as before 1 second of reaching the highest point. But the direction will be changed that is after 1 second the direction will be downward.

So, to sum up, one can say that the velocity after 1 second of reaching the highest point will be:

  $v_2=9.8\text{ m/s}$

Conclusion:

The velocity after 1 sec, $v_2=9.8\text{ m/s}$ .

To determine

Change in velocity during this 1 sec interval.

Answer to Problem 55A

Change in the velocity during this one second interval is $9.8\text{ m/s}$ .

Explanation of Solution

Given:

Final velocity, $v=0$

Velocity before 1 sec, $v_1=9.8\text{ m/s}$

Velocity after 1 sec, $v_2=9.8\text{ m/s}$

Calculation:

Change in the velocity during this time interval will be:

  $\begin{array}{c}\Delta v=v_2-v\\ \Delta v=9.8-0\\ =9.8\text{ m/s}\end{array}$

Conclusion:

Change in the velocity during this one second interval is $9.8\text{ m/s}$ .

To determine

Change in velocity during 2 sec interval.

Answer to Problem 55A

Change in the velocity during 2 second interval is $20\text{ m/s}$ .

Explanation of Solution

Given:

Final velocity, $v=0$

Velocity before 1 sec, $v_1=9.8\text{ m/s}$

Velocity after 1 sec, $v_2=9.8\text{ m/s}$

Calculation:

Change in the velocity during 2 second interval is given by:

  $\begin{array}{c}\Delta v=v_1+v_2\\ \Delta v=9.8+9.8\\ =20\text{ m/s}\end{array}$

Conclusion:

Change in the velocity during 2 second interval is $20\text{ m/s}$ .

To determine

Acceleration of the ball.

Answer to Problem 55A

The acceleration of the ball, $a=9.8{\text{ m/s}}^2$ .

Explanation of Solution

Given:

Final velocity, $v=0$

Velocity before 1 sec, $v_1=9.8\text{ m/s}$

Velocity after 1 sec, $v_2=9.8\text{ m/s}$

Acceleration due to gravity, $g=10{\text{ m/s}}^2$

Formula used:

  $a=\frac{\text{change in velocity}}{t}$

Calculation:

Since, the acceleration is $g=9.8{\text{ m/s}}^2$ , during the whole time, whether moving upward or downward.

Also, using the velocity

  $a=\frac{\text{9}\text{.8 m/s}}{1\text{ s}}=9.8{\text{ m/s}}^2$

Conclusion:

Hence, the acceleration of the ball, $a=9.8{\text{ m/s}}^2$ .