Two bodies are projected from the same point with equal speeds in such directions that they both strikes the same point on a plane whose inclination is $\beta .$ If a be the angle of projection of the first body with the horizontal, the ratio of their times of flight is

See fig. (3) 

Let α be the angle of projection of the second body. The range of both the bodies will be the same. Therefore,

$\begin{array}{l}\mathrm{R}=\frac{{\mathrm{u}}^2}{{\mathrm{gcos}}^2\mathrm{\beta }}[\sin (2\mathrm{\alpha }-\mathrm{\beta })-\sin \mathrm{\beta }]\\ \text{ or }\sin (2\mathrm{\alpha }-\mathrm{\beta })=\sin \left(2{\mathrm{\alpha }}^{\mathrm{\prime }}-\mathrm{\beta }\right)\\ \text{ or }\begin{array}{l}\left(2{\mathrm{\alpha }}^{\mathrm{\prime }}-\mathrm{\beta }\right)=\mathrm{\pi }=(2\mathrm{\alpha }-\mathrm{\beta })\\ \text{ Now, }\mathrm{T}=\frac{2\mathrm{usin}(\mathrm{\alpha }-\mathrm{\beta })}{\mathrm{gcos}\mathrm{\beta }}\text{ and }{\mathrm{T}}^{\mathrm{\prime }}=\frac{2\mathrm{usin}\left({\mathrm{\alpha }}^{\mathrm{\prime }}-\mathrm{\beta }\right)}{\mathrm{gcos}\mathrm{\beta }}\\ \therefore \frac{\mathrm{T}}{{\mathrm{T}}^{\mathrm{\prime }}}=\frac{\sin (\mathrm{\alpha }-\mathrm{\beta })}{\sin \left({\mathrm{\alpha }}^{\mathrm{\prime }}-\mathrm{\beta }\right)}=\frac{\sin (\mathrm{\alpha }-\mathrm{\beta })}{\sin \left[\frac{\mathrm{\pi }}{2}-(\mathrm{\alpha }-\mathrm{\beta })-\mathrm{\beta }\right]}\\ =\frac{\sin (\mathrm{\alpha }-\mathrm{\beta })}{\sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }\right)}=\frac{\sin (\mathrm{\alpha }-\mathrm{\beta })}{\cos \mathrm{\alpha }}\end{array}\end{array}$