Chapter 3, Problem 2SP

Two balls are released simultaneously from the top of a tall building. Ball A is simply dropped with no initial velocity, and ball B is thrown downward with an initial velocity of 15 m/s.

1. a. What are the velocities of the two balls 1.3 seconds after they are released?
2. b. How far has each ball dropped in 1.3 seconds?
3. c. Does the difference in the velocities of the two balls change at any time after their release? Explain.

To determine

The velocities of the two balls $1.3\text{s}$ after they are released.

Answer to Problem 2SP

The velocities of the two balls A and B $1.3\text{s}$ after they are released are $13\text{m/s}$ and $28\text{m/s}$ respectively.

Explanation of Solution

Given Info: A ball is thrown downward with initial velocity $15\text{m/s}$ and another is dropped with no initial velocity.

Write the expression for the final velocity.

$v=v_0+gt$

Here,

$v$ is the final velocity

$v_0$ is the initial velocity

$t$ is the time

$g$ is the acceleration due to gravity

For the ball with the initial velocity, substitute $15\text{m/s}$ for $v_0$, $10{\text{m/s}}^2$ for $g$ and $1.3\text{s}$ for $t$ to get $v$.

$\begin{array}{c}v=\left(15\text{m/s}\right)+\left(10{\text{m/s}}^2\right)\left(1.3\text{s}\right)\\ =28\text{m/s}\end{array}$

For the ball starting from rest, substitute $0\text{m/s}$ for $v_0$, $10{\text{m/s}}^2$ for $g$ and $1.3\text{s}$ for $t$ to get $v$.

$\begin{array}{c}v=\left(0\text{m/s}\right)+\left(10{\text{m/s}}^2\right)\left(1.3\text{s}\right)\\ =13\text{m/s}\end{array}$

Conclusion:

Thus, the velocities of the two balls A and B $1.3\text{s}$ after they are released are $13\text{m/s}$ and $28\text{m/s}$ respectively.

To determine

The distance each ball has dropped in $1.3\text{s}$.

Answer to Problem 2SP

The distance each ball A and B has dropped in $1.3\text{s}$ are $8.45\text{m}$ and $11.05\text{m}$ respectively.

Explanation of Solution

Given Info: A ball is thrown downward with initial velocity $15\text{m/s}$ and another is dropped with no initial velocity.

Write the expression for the distance travelled.

$d=v_{0}t+\frac{1}{2}g{t}^2$

Here,

$d$ is the distance travelled

For the ball with the initial velocity, substitute $15\text{m/s}$ for $v_0$, $10{\text{m/s}}^2$ for $g$ and $1.3\text{s}$ for $t$ to get $d$.

$\begin{array}{c}d=\left(15\text{m/s}\right)\left(1.3\text{s}\right)+\frac{1}{2}\left(10{\text{m/s}}^2\right){\left(1.3\text{s}\right)}^2\\ =11.05\text{m}\end{array}$

For the ball starting from rest, substitute $0\text{m/s}$ for $v_0$, $10{\text{m/s}}^2$ for $g$ and $1.3\text{s}$ for $t$ to get $d$.

$\begin{array}{c}d=\left(0\text{m/s}\right)\left(1.3\text{s}\right)+\frac{1}{2}\left(10{\text{m/s}}^2\right){\left(1.3\text{s}\right)}^2\\ =8.45\text{m}\end{array}$

Conclusion:

Thus, the distance each ball A and B has dropped in $1.3\text{s}$ are $8.45\text{m}$ and $11.05\text{m}$ respectively.

To determine

Whether the difference in the velocities of the two balls change at any time.

Answer to Problem 2SP

The difference in the velocities of the two balls does not change at any time as it is given by the initial velocity of the balls.

Explanation of Solution

Write the expression for the final velocity.

$v=v_0+gt$

This equation gives the final velocity of the balls. The difference in the velocity is given by the initial velocity as the change is usually the acceleration due to gravity multiplied by the time which is the same for both the balls. Thus the difference is only in the initial velocity of the ball which is also a constant value and therefore, does not change at any time.

Conclusion:

Thus, the difference in the velocities of the two balls does not change at any time as it is given by the initial velocity of the balls.