Question

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

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

Answer 1

Answer: (C)

400 m/s

Answer 2

We know that in projection of the particle, for maximum range, $\theta=45^{\circ}$
Now maximum range
$R=\frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2} \sin 90^{\circ}}{g}$
$u=\sqrt{R g}\,\,\,\,...(1)$
Here $: R_{\max }=16\, km =16 \times 10^{3} \,m$,
$g=10 \,m / s ^{2}$
Now from e (1), we get
$u=\sqrt{16 \times 10^{3} \times 10}=400 \,m / s$