question_answer

Two projectiles are fired at different angles with the same magnitude of velocity, such that they have the same range. At what angles they might have been projected?

A) \[25{}^\circ \text{ }and\text{ }65{}^\circ \]          

B) \[35{}^\circ \text{ }and\text{ }75{}^\circ \]

C) \[10{}^\circ \text{ }and\text{ }50{}^\circ \]          

D)  None of these

Correct Answer: A

Solution :

                Range of a projectile (R) is given by \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] where\[\mu \]is velocity of projection and 6 the angle of projection. Substituting\[(90{}^\circ -\theta )\]in place of\[\theta \], we get \[R=\frac{{{u}^{2}}\sin 2({{90}^{o}}-\theta )}{g}\] \[=\frac{{{u}^{2}}\sin ({{180}^{o}}-2\theta )}{g}=\frac{{{u}^{2}}\sin 2\theta }{g}\] Hence, horizontal range is same whether the body is projected at\[\theta \]or\[{{90}^{o}}-\theta \]. \[\therefore \]Therefore,\[{{25}^{o}}\]or\[({{90}^{o}}-25){}^\circ =65{}^\circ \]