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)  \[{{25}^{o}}\] and \[{{65}^{o}}\]   

B)  \[{{35}^{o}}\] and \[{{75}^{o}}\]

C)  \[{{10}^{o}}\] and \[{{50}^{o}}\]

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 9 the angle of projection. Substituting \[({{90}^{o}}-\theta )\] in place of \[\theta ,\] we get \[R=\frac{{{u}^{2}}\sin \,\,({{90}^{o}}-\theta )}{g}\] \[=\frac{{{u}^{2}}\sin ({{180}^{o}}-2\theta )}{g}=\frac{{{u}^{2}}\,\sin \,2\theta }{g}\] Hence horizontal range is same whether the body is projected at \[\theta \] or \[{{90}^{o}}-\theta \]. \[\therefore \] Therefore, \[{{25}^{o}}\] or \[{{({{90}^{o}}-25)}^{o}}={{65}^{o}}\]