Q93 / 160 · JIPMER 2005 · Physics

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?

  • A
    $200 \mathrm{~m} / \mathrm{s}$
  • B
    $100 \mathrm{~m} / \mathrm{s}$
  • Correct answer
    C
    $400 \mathrm{~m} / \mathrm{s}$
  • D
    $300 \mathrm{~m} / \mathrm{s}$

Solution

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 $u=\sqrt{16 \times 10^3 \times 10}=400 \mathrm{~m} / \mathrm{s}$

Hence, the correct answer is option: 'C'.

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