A projectile is thrown in the upward direction making an angle of 60° with the horizontal direction with a velocity of 150 ms^-1. Then the time after which its inclination with the horizontal is 45° is

At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then
$$\begin{gathered} 150\times \frac{1}{2}=\mathrm{v}\times \frac{1}{\sqrt{2}} \text{ or } \mathrm{v}=\frac{150}{\sqrt{2}}{ \mathrm{ms}}^{-1} \\ \text{ Initially: }{\mathrm{u}}_{\mathrm{y}}=\mathrm{usin}60°=\frac{150\sqrt{3}}{2}{ \mathrm{ms}}^{-1} \\ \text{ Finally: }{\mathrm{v}}_{\mathrm{y}}=\mathrm{vsin}45°=\frac{150}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{150}{2}{ \mathrm{ms}}^{-1} \\ \text{ But }\mathrm{vy}={\mathrm{uy}}_{\mathrm{y}}+\mathrm{ayt} \text{ or } \frac{150}{2}=\frac{150\sqrt{3}}{2}-10\mathrm{t}\mid \\ 10\mathrm{t}=\frac{150}{2}(\sqrt{3}-1) \text{ or } \mathrm{t}=7.5(\sqrt{3}-1) \end{gathered}$$