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 \[{{H}_{2}}=\frac{u_{2}^{2}{{\cos }^{2}}\theta }{2g}\] \[\therefore \] \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{u_{1}^{2}}{u_{2}^{2}{{\cos }^{2}}\theta }=\frac{u_{2}^{2}{{\cos }^{2}}\theta }{u_{2}^{2}{{\cos }^{2}}\theta }=\mathbf{1:1}\]
You need to login to perform this action.
You will be redirected in
3 sec