The direction of projectile at certain instant is inclined at angle $\mathrm{\alpha }$ to the horizontal. After t sec, if it is inclined at an angle $\mathrm{\beta }$ then the horizontal component of velocity is

Vertical Component of velocity at time t is $\mathrm{usin\theta }-{\mathrm{gt}}_1$

Horizontal Component of velocity at time t is $\mathrm{ucos\theta }$

$\begin{array}{l}\Rightarrow \tan \mathrm{\alpha }=\frac{\mathrm{usin}\mathrm{\theta }-{\mathrm{gt}}_1}{\mathrm{ucos}\mathrm{\theta }}\\ \& \tan \mathrm{\beta }=\frac{\mathrm{usin}\mathrm{\theta }-\mathrm{g}\left(\mathrm{t}+{\mathrm{t}}_1\right)}{\mathrm{ucos}\mathrm{\theta }}\\ \Rightarrow \mathrm{usin}\mathrm{\theta }-{\mathrm{gt}}_1=\mathrm{ucos}\mathrm{\theta tan}\mathrm{\alpha }\\ \& \mathrm{usin}\mathrm{\theta }-{\mathrm{gt}}_1-\mathrm{gt}=\mathrm{ucos}\mathrm{\theta tan}\mathrm{\beta }\end{array}$

$\begin{array}{l}\text{⇒ }\mathrm{gt}=\mathrm{ucos}\mathrm{\theta }[\tan \mathrm{\alpha }-\tan \mathrm{\beta }]\\ \mathrm{ucos}\mathrm{\theta }=\frac{\mathrm{gt}}{\tan \mathrm{\alpha }-\tan \mathrm{\beta }}\end{array}$