Question

The maximum range for a shell fired by a cannon on horizontal terrain is $20\text{km}$. What must be the muzzle velocity of the shell? (Take $g=10{\text{m/s}}^2$)

• A. $225\text{m/s}$
• B. $126\text{m/s}$
• C. $150\sqrt{3}\text{m/s}$
• D. $200\sqrt{5}\text{m/s}$

Answer 1

The correct option is D $200\sqrt{5}\text{m/s}$

Formula for range is,
$R=\frac{u^2\sin 2\theta }{g}$

Let the muzzle velocity be $u$.
For maximum range, numerator should be maximized.
Since speed for fired shell is equal to $u\text{m/s}$ every time. ($u$ constant)
Maximum range will correspond to maximum value of $\sin 2\theta$, which is 1.
$\sin 2\theta =1\Rightarrow 2\theta ={90}^{\circ }$
Thus, $\theta ={45}^{\circ }$ should be the angle of firing (angle of projection) for shell to obtain maximum range.

$R_{\text{max}}=\frac{u^2}{g}$
$\Rightarrow 20\times {10}^3=\frac{u^2}{10}$
$\therefore u=\sqrt{20\times {10}^4}=200\sqrt{5}\text{m/s}$