Search any question & find its solution
Question:
Answered & Verified by Expert
The maximum range of a gun on horizontal terrain is 16 km , if $g=10 \mathrm{~m} / \mathrm{s}^2$. What must be the muzzle velocity of the shell?
Options:
Solution:
1398 Upvotes
Verified Answer
The correct answer is:
$400 \mathrm{~m} / \mathrm{s}$
We know that in projection of the particle, for maximum range, $\theta=45^{\circ}$
Now maximum range
$\begin{aligned} & R=\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \sin 90^{\circ}}{g} \\ & u=\sqrt{R g} & ...(1) \end{aligned}$
Here : $R_{\max }=16 \mathrm{~km}=16 \times 10^3 \mathrm{~m}$,
$g=10 \mathrm{~m} / \mathrm{s}^2$
Now from eq. (1), we get
$u=\sqrt{16 \times 10^3 \times 10}=400 \mathrm{~m} / \mathrm{s}$
Now maximum range
$\begin{aligned} & R=\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \sin 90^{\circ}}{g} \\ & u=\sqrt{R g} & ...(1) \end{aligned}$
Here : $R_{\max }=16 \mathrm{~km}=16 \times 10^3 \mathrm{~m}$,
$g=10 \mathrm{~m} / \mathrm{s}^2$
Now from eq. (1), we get
$u=\sqrt{16 \times 10^3 \times 10}=400 \mathrm{~m} / \mathrm{s}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.