- Home
- Standard 11
- Physics
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
$2:1$
$1:1$
$2:3$
$1:2$
Solution
$\begin{array}{l}
Horizontal\,range{\kern 1pt} R = \frac{{{u^2}\sin 2\theta }}{g}\\
For\,angle\,of\,projection\,\left( {{{45}^ \circ } – \theta } \right),\,the\\
horizontal\,range\,is\,\\
\therefore \,{R_1} = \frac{{{u^2}\sin \left[ {2\left( {{{45}^ \circ } – \theta } \right)} \right]}}{g}\\
\,\,\,\,\,\,\,\,\,\,\, = \frac{{{u^2}\,\sin \left( {{{90}^ \circ } – 2\theta } \right)}}{g}\\
\,\,\,\,\,\,\,\,\,\,\, = \frac{{{u^2}\cos \,2\theta }}{g}
\end{array}$
$\begin{array}{l}
For\,angle\,of\,projection\,\left( {{{45}^ \circ } + \theta } \right),\,the\\
horizontal\,range\,is\\
{R_2} = \frac{{{u^2}\sin \left[ {2\left( {{{45}^ \circ } + \theta } \right)} \right]}}{g}\\
\,\,\,\,\,\, = \frac{{{u^2}\,\sin \left( {{{90}^ \circ } + 2\theta } \right)}}{g} = \frac{{{u^2}\cos \,2\theta }}{g}\\
\therefore \,\,\,\,\frac{{{R_1}}}{{{R_2}}} = \frac{{{u^2}\cos 2\theta /g}}{{{u^2}\cos 2\theta /g}} = \frac{1}{1}.\\
\therefore \,The\,range\,is\,the\,same.
\end{array}$
Similar Questions
Match the columns
Column $-I$ $R/H_{max}$ |
Column $-II$ Angle of projection $\theta $ |
$A.$ $1$ | $1.$ ${60^o}$ |
$B.$ $4$ | $2.$ ${30^o}$ |
$C.$ $4\sqrt 3$ | $3.$ ${45^o}$ |
$D.$ $\frac {4}{\sqrt 3}$ | $4.$ $tan^{-1}\,4\,=\,{76^o}$ |