**Solution :**

Given :

Area of plate, $\quad A=2 \times 1=2 \mathrm{~m}^{2}$

Velocity of air, $V =50 \mathrm{~km} / \mathrm{hr}
=13.89 \mathrm{~m} / \mathrm{s}
$

Density of air, $\rho =1.15 \mathrm{~kg} / \mathrm{m}^{3}$

Value of $C_{D} =0.15$ and $C_{L}=0.75$

**(i) Lift Force $(F_L)$**

$$\begin{aligned}
F_{L} &=C_{L} \times A \times \rho \times V^{2} / 2 \\
&=0.75 \times 2 \times 1.15 \times 13.89^{2} / 2\\
&=166.404 \mathrm{~N}
\end{aligned}$$

**(ii) Drag force $(F_D)$**

$$\begin{aligned}F_{D} &=C_{D} \times A \times \rho \times U^{2} / 2 \\
&=0.15 \times 2 \times 1.15 \times 13.89^{2} / 2\\
&=33.28 \mathrm{~N}
\end{aligned}$$

**(iii) Resultant force $(F_{R})$**

$$\begin {aligned}
F_{R} &=\sqrt{{F_{D}^{2}}+{F_L}^2} \\
&=\sqrt{{{33.28}^{2}}+{166.404}^2} \\
&=169.67 \mathrm{~N}
\end {aligned}$$

**(iv) The direction of resultant force ( $\theta)$**

The direction of resultant force is given by.

$$
\begin{aligned}
\tan \theta &=\frac{F_{L}}{F_{D}}=\frac{166.38}{33.275}=5 \\
\theta &=\tan ^{-1} 5\\
&=78.69^{\circ}
\end{aligned}
$$

**(v) Power exerted by air on the plate**

Power = Force in the direction of motion × Velocity

$$
\begin{aligned}
&=F_{D} \times U \mathrm{~N} \mathrm{~m} / \mathrm{s} \\
&=33.280 \times 13.89 \mathrm{~W} \\
&=462.26 \mathrm{~W}
\end{aligned}
$$