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A shell is fired from a fixed artillery gun with an initial speed $u$ such that it hits the target on the ground at a distance $R$ from it. If $t_1$ and $t_2$ are the values of the time taken by it to hit the target in two possible ways, the product $t_1t_2$ is
$2R/g$
$R/4g$
$R/g$
$R/2g$
Solution
$\begin{array}{l} Range\,will\,be\,same\,for\,time\,{t_1}\,and\,{t_2},\,so\,\\ angles\,of\,projrection\,will\,be\,'\theta '\,\& '{90^ \circ } – \theta '\\ {t_1} = \frac{{2u\,\sin \,\theta }}{g}{t_2} = \frac{{2u\,\sin \,\left( {{{90}^ \circ } – \theta } \right)}}{g}\,and\\ \,\,\,\,\,\,\,R = \,\frac{{{u^2}\sin \,2\theta }}{g}\\ {t_1}{t_2} = \frac{{4{u^2}\sin \theta \cos \theta }}{{{g^2}}} = \frac{2}{g}\left[ {\frac{{2{u^2}\sin \theta \cos \theta }}{g}} \right]\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2R}}{g} \end{array}$
Similar Questions
Match the columns
| Column $-I$ $R/H_{max}$ |
Column $-II$ Angle of projection $\theta $ |
| $A.$ $1$ | $1.$ ${60^o}$ |
| $B.$ $4$ | $2.$ ${30^o}$ |
| $C.$ $4\sqrt 3$ | $3.$ ${45^o}$ |
| $D.$ $\frac {4}{\sqrt 3}$ | $4.$ $tan^{-1}\,4\,=\,{76^o}$ |
