Chapter 3, Problem 1SP

A ball is thrown straight upward with an initial velocity of 18 m/s. Use g = 10 m/s² for computations listed here.

1. a. What is its velocity at the high point in its motion?
2. b. How much time is required to reach the high point?
3. c. How high above its starting point is the ball at its high point?
4. d. How high above its starting point is the ball 3 seconds after it is released?
5. e. Is the ball moving up or down 2 seconds after it is released?

To determine

The velocity of the ball at the high point.

Answer to Problem 1SP

The velocity of the ball at the high point is $0\text{m/s}$.

Explanation of Solution

Given Info: A ball is thrown up with initial velocity $18\text{m/s}$.

The velocity is defined as the rate of change of displacement of the object.

The ball is thrown up. The displacement is upwards but it is acted upon by the acceleration due to gravity which is acting in the downward direction. This means that the velocity decreases as it goes up. This means that velocity becomes zero as the ball is at the highest point. Therefore, the velocity is $0\text{m/s}$ at the high point.

Conclusion:

Thus, the velocity of the ball at the high point is $0\text{m/s}$.

To determine

The time taken by the ball to reach the high point.

Answer to Problem 1SP

The time taken by the ball to reach the high point is $1.8\text{s}$.

Explanation of Solution

Given Info: A ball is thrown up with initial velocity $18\text{m/s}$.

Write the expression for the final velocity at the high point.

$v=v_0-gt$

Here,

$v$ is the final velocity

$v_0$ is the initial velocity

$g$ is the acceleration due to gravity

$t$ is the time

Substitute $0\text{ m/s}$ for $v$ and re-write in terms of $t$.

$\begin{array}{c}0\text{ m/s}=v_0-gt\\ t=\frac{v_0}{g}\end{array}$

Substitute $18\text{m/s}$ for $v_0$ and $10{\text{m/s}}^2$ for $g$ to get $t$.

$\begin{array}{c}t=\frac{18\text{m/s}}{10{\text{m/s}}^2}\\ =1.8\text{s}\end{array}$

Conclusion:

Thus, the time taken by the ball to reach the high point is $1.8\text{s}$.

To determine

The height of the ball above its starting point at its high point.

Answer to Problem 1SP

The height of the ball above its starting point at its high point is $16.2\text{m}$.

Explanation of Solution

Given Info: A ball is thrown up with initial velocity $18\text{m/s}$.

Write the expression for the vertical distance travelled.

$d_v=v_{0}t-\frac{1}{2}g{t}^2$

Here,

$d_v$ is the vertical distance travelled

Substitute $18\text{m/s}$ for $v_0$, $1.8\text{s}$ for $t$ and $10{\text{m/s}}^2$ for $g$ to get $d_h$.

$\begin{array}{c}{d}_v=\left(18\text{m/s}\right)\left(1.8\text{s}\right)-\frac{1}{2}\left(10{\text{m/s}}^2\right){\left(1.8\text{s}\right)}^2\\ =16.2\text{m}\end{array}$

Conclusion:

Thus, the height of the ball above its starting point at its high point is $16.2\text{m}$.

To determine

The height of the ball above its starting point $3\text{s}$ after the release.

Answer to Problem 1SP

The height of the ball above its starting point $3\text{s}$ after the release is $9\text{m}$.

Explanation of Solution

Given Info: A ball is thrown up with initial velocity $18\text{m/s}$.

Write the expression for the vertical distance travelled.

$d_v=v_{0}t-\frac{1}{2}g{t}^2$

Substitute $18\text{m/s}$ for $v_0$, $3\text{s}$ for $t$ and $10{\text{m/s}}^2$ for $g$ to get $d_h$.

$\begin{array}{c}{d}_v=\left(18\text{m/s}\right)\left(3\text{s}\right)-\frac{1}{2}\left(10{\text{m/s}}^2\right){\left(3\text{s}\right)}^2\\ =9\text{m}\end{array}$

Conclusion:

Thus, the height of the ball above its starting point $3\text{s}$ after the release is $9\text{m}$.

To determine

Whether the ball is moving up or down $2\text{s}$ after the release.

Answer to Problem 1SP

The ball is moving down $2\text{s}$ after the release.

Explanation of Solution

Given Info: A ball is thrown up with initial velocity $18\text{m/s}$.

Write the expression for the final velocity at the high point.

$v=v_0-gt$

Substitute $0\text{ m/s}$ for $v$ and re-write in terms of $t$.

$\begin{array}{c}0\text{ m/s}=v_0-gt\\ t=\frac{v_0}{g}\end{array}$

Substitute $18\text{m/s}$ for $v_0$ and $10{\text{m/s}}^2$ for $g$ to get $t$.

$\begin{array}{c}t=\frac{18\text{m/s}}{10{\text{m/s}}^2}\\ =1.8\text{s}\end{array}$

This is less than $2\text{s}$. This means that the ball has already reached the high point and is moving down at $2\text{s}$.

Conclusion:

Thus, the ball is moving down $2\text{s}$ after the release.