JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 21)
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 t1 and t2 are the values of the time taken by it to hit the target in two possible ways, the
product t1t2 is -
$${{2R} \over g}$$
$${R \over g}$$
$${R \over {2g}}$$
$${R \over {4g}}$$
Explanation
Range will be same for time t1 and t2, so angles of projection will be ‘$$\theta $$’ & ‘90° – $$\theta $$’
$${t_1} = {{2u\sin \theta } \over g}{t_2} = {{2u\sin \left( {{{90}^o} - \theta } \right)} \over g}$$
and $$R = {{{u^2}\sin 2\theta } \over g}$$
$${t_1}{t_2} = {{4{u^2}\sin \theta \cos \theta } \over {{g^2}}} = {2 \over g}\left[ {{{2{u^2}\sin \theta \cos \theta } \over g}} \right]$$
= $${{2R} \over g}$$
$${t_1} = {{2u\sin \theta } \over g}{t_2} = {{2u\sin \left( {{{90}^o} - \theta } \right)} \over g}$$
and $$R = {{{u^2}\sin 2\theta } \over g}$$
$${t_1}{t_2} = {{4{u^2}\sin \theta \cos \theta } \over {{g^2}}} = {2 \over g}\left[ {{{2{u^2}\sin \theta \cos \theta } \over g}} \right]$$
= $${{2R} \over g}$$
Comments (0)
