question_answer

A ball of mass m is thrown vertically up. Another ball of mass 2m is thrown at an angle \[\theta \] with the vertical. Both of them remain in air for the same period of time. What is the ratio of the heights attained by the two balls?

Answer:

                Time of flight of the ball thrown vertically upwards, \[{{T}_{1}}=\frac{2{{u}_{1}}}{g}\] Time of flight of the ball thrown at an angle \[\theta \] with the vertical , \[{{T}_{\,}}_{2}=\frac{2{{u}_{2}}\cos \theta }{g}\] As                           \[{{T}_{1}}={{T}_{2}}\] \[\therefore \]  \[\frac{2{{u}_{1}}}{g}=\frac{2{{u}_{2}}\cos \theta }{g}\] or            \[{{u}_{1}}={{u}_{2}}\cos \theta \] Now      \[{{H}_{}}_{1}=\frac{u_{1}^{2}}{2g}\]      and \[{{H}_{2}}=\frac{u_{2}^{2}{{\cos }^{2}}\theta }{2g}\] \[\therefore \]  \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{u_{1}^{2}}{u_{2}^{2}{{\cos }^{2}}\theta }=\frac{u_{2}^{2}{{\cos }^{2}}\theta }{u_{2}^{2}{{\cos }^{2}}\theta }=\mathbf{1:1}\]