A cricket ball thrown across a field is at heights $${h_1}$$ and $${h_2}$$ from point of projection at times $${t_1}$$ and $${t_2}$$ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

Solution :
$$\eqalign{ & {h_1} = u\sin \theta {t_1} + \frac{1}{2}gt_1^2; \cr & {h_2} = u\sin \theta {t_2} + \frac{1}{2}gt_2^2 \cr & {\text{So,}}\,\,\frac{{{t_1}}}{{{t_2}}} = \frac{{{h_1} + \frac{1}{2}gt_1^2}}{{{h_2} + \frac{1}{2}gt_2^2}} \cr & \Rightarrow {h_1}{t_2} - {h_2}{t_1} = \frac{1}{2}g\left( {{t_1}t_2^2 - t_1^2{t_2}} \right) \cr & {\text{Time of flight}} = \frac{{2u\sin \theta }}{g} = \frac{{{h_1}t_2^2 - {h_2}t_1^2}}{{{h_1}{t_2} - {h_2}{t_1}}}\,\left[ {{\text{Use above eqn to simplify}}} \right] \cr} $$