Question

The aeroplane is shown here on level flight at an altitude of $0.5\mathrm{km}$ and at a speed of $150{\mathrm{kmh}}^{-1}$.

At what distance $x$ should it release a bomb to hit the target? [Take $g=10{\mathrm{ms}}^{-2}$]

• A. $150\mathrm{m}$
• B. $420\mathrm{m}$
• C. $2000\mathrm{m}$
• D. $800\mathrm{m}$

Answer 1

The correct option is B

$420\mathrm{m}$

Step 1: Given data

The height at which the aeroplane is flying, $H=0.5km=500m$

Speed of the aeroplane, $u=150km{h}^{-1}=\frac{150\times 5}{18}=\frac{125}{3}m{s}^{-1}$

The distance between the aeroplane and the target is $x$

Acceleration due to gravity, $g=10{\mathrm{ms}}^{-2}$

The diagrammatic representation of the given data is given below:

Step 2: Find the distance from the target where the aeroplane should release the bomb

As the aeroplane is moving with a velocity of $\frac{125}{3}m{s}^{-1}$that is why when the bomb will be thrown it will also have an initial velocity equal to the velocity of the aeroplane. Hence, the motion of the bomb while falling will be a projectile motion.

$\mathbf{\therefore }\mathbf{}\mathit{H}\mathbf{=}\frac{\mathbf{g}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{u}}^{\mathbf{2}}}$

$$\begin{gathered} \Rightarrow 500=\frac{10\times x^2}{2\times {\left({\displaystyle \frac{125}{3}}\right)}^2} \\ \Rightarrow x^2=\frac{500\times 2\times 125\times 125}{3\times 3\times 10} \\ \Rightarrow x=\frac{10\times 125}{3} \\ \Rightarrow x=416.7m=420m\left(approximately\right) \end{gathered}$$

Hence, the distance at which it should release a bomb to hit the target is $\mathbf{420}\mathbf{}\mathit{m}$.