Chapter 2, Problem 17PEB

An object is observed to fall from a bridge, striking the water below 2.50 s later, (a) With what velocity did it strike the water? (b) What was its average velocity during the fall? (c) How high is the bridge?

To determine

The velocity with which the object falling from a bridge strikes the surface of the water after $2.50\text{ s}$.

Answer to Problem 17PEB

Solution:

The velocity with which the object strikes water is $25\text{ m/s}$.

Explanation of Solution

Given data:

The time of travel of water from the bridge to the surface of water is $2.50\text{ s}$.

Formula used:

Write the expression for average acceleration.

$a=\frac{\Delta v}{t}$

Here, $\Delta v$ is the change in velocity and $t$ is the total time taken to reach the final speed.

Write the expression for the change in velocity.

$\Delta v=v_{\text{f}}-v_{\text{i}}$

Here, $v_{\text{f}}$ and $v_{\text{i}}$ are the final and initial velocity.

Explanation:

Refer to the expressions for average acceleration and change in velocity. Combine both the expressions and write the expression for average acceleration in terms of initial and final velocity.

$\begin{array}{c}a=\frac{\Delta v}{t}\\ =\frac{v_{\text{f}}-v_{\text{i}}}{t}\end{array}$ …… (1)

Consider the motion of the object from the bridge to the surface of the water. The object falls from the bridge and strikes the surface of the water after $2.50\text{ s}$. Since the object is initially at rest at the bridge before falling, therefore, its initial velocity is zero. Also, the acceleration acting on the object is due to gravity. Thus, calculate the velocity with which the object strikes the surface of the water.

Substitute $0\text{ m/s}$ for $v_{\text{i}}$, $2.50\text{ s}$ for $t$, and $-9.8{\text{ m/s}}^2$ for $a$ in the equation (1).

$\begin{array}{c}-9.8{\text{ m/s}}^2=\frac{v_{\text{f}}-0\text{ m/s}}{2.50\text{ s}}\\ =\frac{v_{\text{f}}}{2.5\text{ s}}\end{array}$

Further, solve.

$\begin{array}{c}{v}_{\text{f}}=-\left(9.8{\text{ m/s}}^2\right)\left(2.5\text{ s}\right)\\ \approx -25\text{ m/s}\end{array}$

Here, since the gravitational acceleration always acts in the downward direction, therefore, the sign of acceleration is negative. Also, the negative value of final velocity denotes the final velocity is in the downward direction.

Conclusion:

The magnitude of final velocity with which the object strikes water is $25\text{ m/s}$.

To determine

The average velocity of the object during its fall.

Answer to Problem 17PEB

Solution:

The object’s average velocity during the fall is $12.5\text{ m/s}$.

Explanation of Solution

Given data:

The magnitude of initial velocity and final velocity of the object are $0\text{ m/s}$ and $25\text{ m/s}$ respectively.

Formula used:

Write the expression for the average velocity.

$\overline{v}=\frac{v_{\text{i}}+v_{\text{f}}}{2}$

Explanation:

The magnitude of initial velocity and the final velocity of the object during the fall are $0\text{ m/s}$ and $25\text{ m/s}$ respectively. Calculate the average velocity of the object during its fall.

${\left(\overline{v}\right)}_{\text{during fall}}=\frac{v_{\text{i}}+v_{\text{f}}}{2}$

Substitute $0\text{ m/s}$ for $v_{\text{i}}$, and $25\text{ m/s}$ for $v_{\text{f}}$.

$\begin{array}{c}{\left(\overline{v}\right)}_{\text{during fall}}=\frac{0\text{ m/s}+25\text{ m/s}}{2}\\ =12.5\text{ m/s}\end{array}$

Here, only magnitude is taken because direction is not necessary to calculate average velocity. The negative value of final velocity denotes its direction but to find the average velocity, only the magnitude of velocity is needed and thus, the negative sign is not taken.

Conclusion:

The average velocity of the object during the fall is $12.5\text{ m/s}$.

To determine

The height of the bridge from the water surface.

Answer to Problem 17PEB

Solution:

The bridge is at a height of $32\text{ m}$.

Explanation of Solution

Given data:

The average velocity of the object during its fall is $12.5\text{ m/s}$. Also, the time taken by the object to fall from bridge to the water is $2.50\text{ s}$.

Formula used:

Write the expression for the average velocity in terms of distance and time.

$\overline{v}=\frac{d}{t}$

Here, $d$ is the total distance travelled.

Explanation:

The average velocity during the fall of the object from the bridge to water is $12.5\text{ m/s}$. Also, the total time taken during its fall is $2.50\text{ s}$. Calculate the total distance travelled by the object during its fall.

${\left(\overline{v}\right)}_{\text{during fall}}=\frac{d_{\text{bridge}}}{t}$

Substitute $12.5\text{ m/s}$ for ${\left(\overline{v}\right)}_{\text{during fall}}$, and $2.50\text{ s}$ for $t$.

$12.5\text{ m/s}=\frac{d_{\text{bridge}}}{2.50\text{ s}}$

Further, solve for $d_{\text{bridge}}$.

$\begin{array}{c}{d}_{\text{bridge}}=\left(12.5\text{ m/s}\right)\left(2.50\text{ s}\right)\\ \approx 32\text{ m}\end{array}$

Conclusion:

The height of the bridge is $32\text{ m}$.