A ball is projected from the ground at angle $\mathrm{\theta }$ with the horizontal. After 1s it is moving at angle 45° with the horizontal and after 2 s it is moving horizontally. What is the velocity of projection of the ball?

After one second it is moving 45⁰ with horizontal 

$$\begin{gathered} \tan {45}^0=\frac{\mathrm{usin\theta }-\mathrm{g}\times 1}{\mathrm{ucos\theta }}=1 \\ \mathrm{ucos\theta }=\mathrm{usin\theta }-\mathrm{g}\Rightarrow \Rightarrow \left(1\right) \\ \mathrm{after} \mathrm{two} \mathrm{seconds} \mathrm{it} \mathrm{moves} \mathrm{horizontally} \mathrm{means} \mathrm{it} \mathrm{is} \mathrm{at} \mathrm{the} \mathrm{higest} \mathrm{point} \\ \frac{\mathrm{usin\theta }}{\mathrm{g}}=2 \\ \mathrm{usin\theta }=2\mathrm{g} \mathrm{substituting} \mathrm{in} \left(1\right) \\ \mathrm{ucos\theta }=\mathrm{g} \\ \mathrm{u}=\sqrt{{\left(\mathrm{usin\theta }\right)}^2+{\left(\mathrm{ucos\theta }\right)}^2} \\ \mathrm{u}=\sqrt{{\left(2\mathrm{g}\right)}^2+{\left(\mathrm{g}\right)}^2}=\sqrt{5}\mathrm{g}=10\sqrt{5} \end{gathered}$$