JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2014
- Hard
- MCQ
The initial speed of a bullet fired from a rifle is \(630\, m/s\). The rifle is fired at the centre of a target \(700\, m\) away at the same level as the target. How far above the center of the target (in \(m\)) the rifle must be aimed in order to hit the target? (Take \(g=10 \;m/s^2\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(6.17\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
Let\,'t'\,be\,the\,\,time\,taken\,by\,the\,bullet\,\,to\,hit\,the\,{\rm{target}}{\rm{.}}\\
\therefore \,{\rm{700 = }}\,{\rm{m = 63}}{{\rm{0}}^{ - 1}}t\\
\Rightarrow t = \frac{{700m}}{{630m{s^{ - 1}}}} = \frac{{10}}{9}\sec \\
For\,velrtical\,motion,\\
Here,\,u = 0\\
\therefore \,h = \frac{1}{2}g{t^2}
\end{array}\)
Let\,'t'\,be\,the\,\,time\,taken\,by\,the\,bullet\,\,to\,hit\,the\,{\rm{target}}{\rm{.}}\\
\therefore \,{\rm{700 = }}\,{\rm{m = 63}}{{\rm{0}}^{ - 1}}t\\
\Rightarrow t = \frac{{700m}}{{630m{s^{ - 1}}}} = \frac{{10}}{9}\sec \\
For\,velrtical\,motion,\\
Here,\,u = 0\\
\therefore \,h = \frac{1}{2}g{t^2}
\end{array}\)
\(\begin{array}{l}
= \frac{1}{2} \times 10 \times {\left( {\frac{{10}}{9}} \right)^2}\\
= \frac{{500}}{{81}}m = 6.17\,m\\
Therefore,\,the\,rifle\,must\,be\,{\rm{aimed}}\,{\rm{6}}{\rm{.17}}\,{\rm{m}}\,{\rm{above}}\,{\rm{the}}\,{\rm{center}}\,{\rm{of}}\,{\rm{the}}\,\\
{\rm{target}}\,{\rm{to}}\,{\rm{hit}}\,{\rm{the}}\,{\rm{target}}{\rm{.}}\,
\end{array}\)
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