A ball is projected from the ground at an angle $\theta$  with the horizontal. After 1 second it is moving at an angle ${45}^0$  with the horizontal and after 2 seconds it is moving horizontally. What is the velocity of projection of the ball?

Suppose the angle made by the instantaneous velocity with the horizontal be $\alpha$ . Then

$\tan \alpha =\frac{v_y}{v_x}=\frac{u\sin \theta -gt}{u\cos \theta }$

Given that, $\alpha ={45}^0$ , when $t=1\sec$

$\alpha =0^0$ , when $t=2\sec$

This gives, $u\cos \theta =u\sin \theta -g$

$u\sin \theta -2g=0$

Solving we find, $u\sin \theta =2g$

$u\cos \theta =g$

Solving we get, $u=\sqrt{5}$ $g=10\sqrt{5}m{s}^{-1}$