A ball is projected from the ground at angle θ with the horizontal. After 1 s, it is moving at angle 45o with the horizontal and after 2 s 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 α. Then tan⁡α=vyvx=usin⁡θ−gtucos⁡θ Given that α=45∘, when t=1s;α=0∘, when t=2s This gives ucos⁡θ=usin⁡θ−g    ....(i) and usin⁡θ−2g=0   .....(ii) Solving Eqs. (i) and (ii), we find usin⁡θ=2g and ucos⁡θ=g. Squaring and adding,u=5g=105ms−1