Chapter 23, Problem 28SP

A shell fired at a target 800 m away was heard by someone standing near the gun to strike the target 5.0 s after leaving the gun. Compute the average horizontal velocity of the shell. The air temperature is 20 °C.

To determine

The average speed of the shell fired at a target which is $800\text{ m}$ away to strike the target $5\text{s}$ after leaving the gun, when the air temperature is $200°\text{C}$.

Answer to Problem 28SP

Solution:

$0.30\text{ km/s}$

Explanation of Solution

Given data:

Distance of the target from the observer is $800\text{ m}$.

Temperature of the air is $20.0°\text{C}$.

Observer hears the sound of the shell striking the target $5.00\text{ s}$ after the gun is fired.

Formula used:

The relation between temperature and speed of sound is,

$\frac{v}{v_0}=\sqrt{\frac{T}{T_0}}$

Here, $v$ is speed of sound at temperature $T$ and $v_0$ is the speed of sound at temperature $T_0$.

The speed of sound at $0°\text{ C}$ is $331.3\text{ m/s}$.

The formula for distance is

$d=vt$

Here, $d$ is distance, $v$ is speed, and $t$ is time.

Explanation:

Calculate the time that the sound takes to reach the observer from the target when shell hits it.

Recall the formula for unit conversion of temperature.

$T^{\prime }=\left(273+T\right)\text{ K}$

Convert the unit of temperature at $0°\text{ C}$ to kelvin.

Substitute $0°\text{ C}$ for $T$ and $T_0$ for $T^{\prime }$.

$\begin{array}{c}{T}_0=\left(273+0\right)\text{ K}\\ =2\text{73 K}\end{array}$

Convert the unit of temperature at $20.0°\text{C}$ to kelvin.

Substitute $20.0$ for $T$.

$\begin{array}{c}{T}^{\prime }=\left(273+20.0\right)\text{ K}\\ =293\text{ K}\end{array}$

Recall the formula for the speed of sound in terms of temperature,

$\frac{v}{v_0}=\sqrt{\frac{T}{T_0}}$

Rearrange the equation for $v$.

$v=v_0\sqrt{\frac{T}{T_0}}$

Substitute $2\text{73 K}$ for $T_0$, $331.3\text{ m/s}$ for $v_0$, and $293\text{ K}$ for $T^{\prime }$

$\begin{array}{c}v=\left(331.3\text{ m/s}\right)\sqrt{\frac{293\text{ K}}{2\text{73 K}}}\\ =343.2\text{ m/s}\end{array}$

Recall the formula for the distance travelled.

$d=vt$

Rearrange the equation for $t$.

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

Substitute $800\text{ m}$ for $d$ and $343.2\text{ m/s}$ for $v$

$\begin{array}{c}t=\frac{800\text{ m}}{343.2\text{ m/s}}\\ =2.33\text{ s}\end{array}$

Since the observer hears the sound of shell striking the target $5.00\text{ s}$ after the gun is fired. Therefore, time that the shell takes to reach the target is,

$t^{\prime }=\left(5.00\text{ s}\right)-\left(t\right)$

Here, $t^{\prime }$ is the time taken by the shell to reach the target.

Substitute $2.33\text{ s}$ for $t$

$\begin{array}{c}{t}^{\prime }=\left(5.00\text{ s}\right)-\left(2.33\text{ s}\right)\\ =2.67\text{ s}\end{array}$

Now, calculate the speed of the shell.

Recall the formula for the distance travelled.

$d=v^{\prime }{t}^{\prime }$

Here, $v^{\prime }$ is speed of the shell and $t^{\prime }$ is the time that the shell takes to reach the target.

Rearrange the equation for $v$.

$v=\frac{d}{t^{\prime }}$

Substitute $800\text{ m}$ for $d$ and $2.67\text{ s}$ for $t^{\prime }$

$\begin{array}{c}v=\frac{800\text{ m}}{2.67\text{ s}}\\ =299.63\text{ m/s}\left(\frac{1\text{ km/s}}{1\text{000 m/s}}\right)\\ =\text{0}\text{.2996 km/s}\\ =0.30\text{ km/s}\end{array}$

Conclusion:

The speed of the shell is $0.30\text{ km/s}$.