Question

Calculate the magnitude of the drag force on a missile 53 cmin diameter cruising at 250 m/sat low altitude, where the density of air is$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$. Assume$\mathbf{C}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{75}$.

Short answer

The magnitude of the drag force is $D=6.2\times {10}^3\mathrm{N}$.

Step-by-step solution

Given data 

• Radius of missile: $R=\frac{53}{2}\mathrm{cm}=0.265\mathrm{m}$.
• Velocity$=250\mathrm{m}/\mathrm{s}$ .
• The density of air: $\rho =1.2\mathrm{kg}/{\mathrm{m}}^3$.
• Drag coefficient: $C=0.75$.

Understanding the concept

The problem is based on the concept of drag force. In fluid mechanics, it is the force acting opposite to the relative motion of any object moving with respect to a surrounding fluid.

Formula:

$D=\frac{1}{2}C\rho A{v}^2$

Where c = drag coefficient

$\rho$= density of the medium

A = area of cross-section of the body

v = the velocity of the body through the medium

Step 3: Calculate the magnitude of the drag force on a missile

The magnitude of drag force on the missile:

$$\begin{gathered} D=\frac{1}{2}C\rho A{v}^2 \\ =\frac{1}{2}C\rho A\pi R^2{v}^2 \end{gathered}$$

Substitute values in the above expression, and we get,

$$\begin{gathered} D=\frac{1}{2}\times 0.75\times 1.2\mathrm{kg}/{\mathrm{m}}^3\times 3.14\times {\left(0.265\mathrm{m}\right)}^2\times {\left(250\mathrm{m}/\mathrm{s}\right)}^2 \\ D=6201.7\mathrm{N} \end{gathered}$$

Thus, the magnitude of the drag force is$6.2\times {10}^3\mathrm{N}$ .