91Ó°ÊÓ

$\mathrm{Speed}\mathrm{of}\mathrm{the}\mathrm{plane}=1300\mathrm{km}/\mathrm{h}$

This problem is based on the drag force which is a type of friction. This is the force acting opposite to the relative motion of an object moving with respect to the surrounding medium.

Formula:

The terminal speed is given by

$v_t=\sqrt{\frac{2F_g}{CpA}}$

Where C is the drag coefficient,${}^p$is the fluid density, A is the effective cross-sectional area, and${}^{F_g}$is the gravitational force.

From table and equation for the speed of air,

$v_t=\sqrt{\frac{2F_g}{CpA}}$

$CpA=2\frac{mg}{v{t}^2}$

Where,${}^{v_t=60\mathrm{m}/\mathrm{s}}$. Now, estimate the pilot’s mass at about${}^{m=70kg}$

Now, convert,velocity into m/s and plug into equation for drag force

$$\begin{gathered} v=1300\left(1000/3600\right) \\ =360m/s \end{gathered}$$

Thus the drag force is given by,

$$\begin{gathered} D=\frac{1}{2}CpA{v}^2 \\ =\frac{1}{2}\left(2\frac{mg}{{v^2}_t}\right)v^2 \\ =mg{\left(\frac{v}{v_t}\right)}^2 \end{gathered}$$

which yields,

$$\begin{gathered} D=\left(70\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(360/60\right)}^2 \\ ≈2×{10}^4\mathrm{N} \end{gathered}$$

Therefore, the drag force on the pilot seat is${}^{2×{10}^4\mathrm{N}}$

Assume the mass of the ejection seat is roughly equal to the mass of the pilot.Thus, Newton’s second law (in the horizontal direction) applied to this system of mass

2m gives the magnitude of acceleration as

$$\begin{gathered} \left|a\right|=\frac{D}{2m} \\ =\frac{g}{2}{\left(\frac{v}{v_t}\right)}^2 \\ =18g\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the horizontal deceleration is${}^{18\mathrm{g}\mathrm{m}/{\mathrm{s}}^2}$.