(I) (a) How fast is an object moving on land if its speed at 24°C is Mach 0.33? (b) A high-flying jet cruising at 3000 km/h displays a Mach number of 3.1 on a screen. What is the speed of sound at that altitude?

In this problem, the relation of Mach number along with speed of sound will be used to determine the speed of the object moving on the land.

The temperature is\(T = 24^\circ {\rm{C}}\).

The Mach number is\(M = 0.33\).

The speed of the jet is\({v_0} = 3000\;{\rm{km/h}}\).

The Mach number on the screen is \({M_0} = 3.1\).

The relation of Mach number is given by,

\(\begin{aligned}{l}M &= \frac{v}{{{v_{\rm{s}}}}}\\M &= \frac{v}{{\left( {331\;{\rm{m/s}} + 0.6T} \right)}}\end{aligned}\)

Here, \({v_{\rm{s}}}\) is the speed of the sound and \(v\) is the speed of the object.

On plugging the values in the above relation.

\(\begin{aligned}{c}0.33 &= \frac{v}{{\left( {331\;{\rm{m/s}} + 0.6\left( {24^\circ {\rm{C}}} \right)} \right)}}\\v &= 113.98\;{\rm{m/s}}\end{aligned}\)

Thus, \(v = 113.98\;{\rm{m/s}}\) is the required speed.

The relation of Mach number is given by,

\(\begin{aligned}{c}{M_0} &= \frac{{{v_0}}}{{{v_{\rm{s}}}'}}\\{M_0} &= \frac{{{v_0}}}{{{v_{\rm{s}}}'}}\end{aligned}\)

Here, \({v_{\rm{s}}}'\) is the speed of the sound.

On plugging the values in the above relation.

\(\begin{aligned}{c}3.1 &= \left( {\frac{{3000\;{\rm{km/h}} \times \frac{{1\;{\rm{m/s}}}}{{3.6\;{\rm{km/h}}}}}}{{{v_{\rm{s}}}'}}} \right)\\{v_{\rm{s}}}' &= 268.8\;{\rm{m/s}}\end{aligned}\)

Thus, \({v_{\rm{s}}}' = 268.8\;{\rm{m/s}}\) is the required speed of sound.