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Question

Physics Question on Motion in a plane

A rifle shoots a bullet with a muzzle velocity of 500ms1500\,m{{s}^{-1}} at a small target 50 m away. To hit the target the rifle must be aimed: (Take g=10ms2g=10\,m{{s}^{-2}} )

A

exactly at the target

B

10 cm below the target

C

10 cm above the target

D

5 cm above the target

Answer

5 cm above the target

Explanation

Solution

We know that distance
(sd)=speed(v) !!×!! time(t)\text{(sd)}\,\text{=}\,\text{speed}\,\text{(v)}\,\text{ }\\!\\!\times\\!\\!\text{ }\,\text{time}\,\text{(t)}
\Rightarrow t=svt=\frac{s}{v}
Given, v=500ms1,s=50mv=500\,m{{s}^{-1}},s=50\,m
\therefore t=50500=0.1st=\frac{50}{500}=0.1\,s
From equation of motion, for vertical displacement
h=ut+12gt2h=ut+\frac{1}{2}g{{t}^{2}}
Given, u=0,t=0.1su=0,t=0.1\,s
\therefore h=12×10×(0.1)2=0.05m=5cm.h=\frac{1}{2}\times 10\times {{(0.1)}^{2}}=0.05\,m=5\,cm.
Hence, the rifle must be aimed 5 cm above the target.