Question

The maximum range of a gun on horizontal terrain is $16km$ if $g=10m/s^2$ . What must be the muzzle velocity of the shell?

• A. $200m/s$
• B. $100m/s$
• C. $400m/s$
• D. $300m/s$

Answer 1

Answer: (C)

For maximum horizontal range body should be projected at an angle of $45^{\circ }$.
The horizontal range covered is given by $R=u_x\times T$
$=u\cos \theta \times \frac{2u\sin \theta }{g}$
$=u^2\frac{(2\sin \theta \cos \theta )}{g}$
$R=\frac{u^2\sin 20}{g}$

For $R_{\max }\sin 2\theta =1$
$\therefore \theta =45^{\circ }$
$\therefore R_{\max }=\frac{u^2}{g}$
$\Rightarrow u=\sqrt{R_{\max }g}$
Given, $R_{\max }=16km=16\times 10^{3}m,g=10m/s^2$
$u=\sqrt{16\times 10^3\times 10}$
$u=400m/s$
Note: Horizontal range is same whether the body is projected at $\theta$ or $\left(90^{\circ }-\theta \right)$ i. e., for complementary angles of projection range is same.