MCQ
For angles of projection of a projectile at angle $(45^o +\theta)$ and $(45^o -\theta ) $ , the horizontal range described by the projectile are in the ratio of
- A$2:1$
- ✓$1:1$
- C$2:3$
- D$1:2$
✓
Answer
Correct option: B.
$1:1$
b
$\begin{array}{l}
Horizontal\,range{\kern 1pt} R = \frac{{{u^2}\sin 2\theta }}{g}\\
For\,angle\,of\,projection\,\left( {{{45}^ \circ } - \theta } \right),\,the\\
horizontal\,range\,is\,\\
\therefore \,{R_1} = \frac{{{u^2}\sin \left[ {2\left( {{{45}^ \circ } - \theta } \right)} \right]}}{g}\\
\,\,\,\,\,\,\,\,\,\,\, = \frac{{{u^2}\,\sin \left( {{{90}^ \circ } - 2\theta } \right)}}{g}\\
\,\,\,\,\,\,\,\,\,\,\, = \frac{{{u^2}\cos \,2\theta }}{g}
\end{array}$
$\begin{array}{l}
Horizontal\,range{\kern 1pt} R = \frac{{{u^2}\sin 2\theta }}{g}\\
For\,angle\,of\,projection\,\left( {{{45}^ \circ } - \theta } \right),\,the\\
horizontal\,range\,is\,\\
\therefore \,{R_1} = \frac{{{u^2}\sin \left[ {2\left( {{{45}^ \circ } - \theta } \right)} \right]}}{g}\\
\,\,\,\,\,\,\,\,\,\,\, = \frac{{{u^2}\,\sin \left( {{{90}^ \circ } - 2\theta } \right)}}{g}\\
\,\,\,\,\,\,\,\,\,\,\, = \frac{{{u^2}\cos \,2\theta }}{g}
\end{array}$
$\begin{array}{l}
For\,angle\,of\,projection\,\left( {{{45}^ \circ } + \theta } \right),\,the\\
horizontal\,range\,is\\
{R_2} = \frac{{{u^2}\sin \left[ {2\left( {{{45}^ \circ } + \theta } \right)} \right]}}{g}\\
\,\,\,\,\,\, = \frac{{{u^2}\,\sin \left( {{{90}^ \circ } + 2\theta } \right)}}{g} = \frac{{{u^2}\cos \,2\theta }}{g}\\
\therefore \,\,\,\,\frac{{{R_1}}}{{{R_2}}} = \frac{{{u^2}\cos 2\theta /g}}{{{u^2}\cos 2\theta /g}} = \frac{1}{1}.\\
\therefore \,The\,range\,is\,the\,same.
\end{array}$
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