Question

A missile launched from the ground, making an angle of ${30}^o$ with horizontal, reaches the target at a distance of $500m$. What is the initial velocity of the missile?

• A. $\sqrt{\frac{{10}^5}{\sqrt{3}}}$
• B. $\sqrt{\frac{{10}^4}{\sqrt{3}}}$
• C. $\sqrt{\frac{100}{\sqrt{2}}}$
• D. $\sqrt{\frac{{10}^4}{3}}$

Answer 1

The correct option is B

$\sqrt{\frac{{10}^4}{\sqrt{3}}}$

Step 1: Given data and formula used

Range $\mathrm{R}=500\mathrm{m},\mathrm{g}=10{\mathrm{ms}}^{-2},\mathrm{\theta }={30}^{\mathrm{o}}$

Formula used :

$$\begin{gathered} \mathrm{R}=\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}} \\ {\mathrm{u}}^2=\frac{\mathrm{Rg}}{\sin 2\mathrm{\theta }} \end{gathered}$$

Step 2: Calculation of initial velocity of the missile$\left(\mathrm{u}\right)$

On putting the given value in the above formula

$$\begin{gathered} {\mathrm{u}}^2=\frac{10\times 500}{\sin \left({60}^0\right)}=\frac{5000}{1}\times \frac{2}{\sqrt{3}}=\frac{{10}^4}{\sqrt{3}} \\ \mathrm{u}=\sqrt{\frac{{10}^4}{\sqrt{3}}} \end{gathered}$$

Hence, the correct option is (b).