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?

Correct answer: C. 400 ~m / s We know that in projection of the particle, for maximum range, θ=45^ Now maximum range aligned & R=(u^2 2 θ)/(g)= u^2 90^ g & u= R g & ...(1) aligned Here : R_ =16 ~km=16 × 10^3 ~m , g=10 ~m / s^2 Now from eq. (1), we get u= 16 × 10^3 × 10 =400 ~m / s

JIPMER 2005

Exam: JIPMER
Year: 2005
Subject: Physics
Chapter: Motion in Two Dimensions

JIPMER 2005 · Physics · Motion in Two Dimensions

The maximum range of a gun on horizontal terrain is 16 km , if g=10...

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

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

Correct answer

C. $400 \mathrm{~m} / \mathrm{s}$

Step-by-step 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}

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