Question

A ball is projected from the ground at angle tita with the horizontal.After 1 sec it is moving at angle 45° with the horizontal and after 2s it is moving horizontally.What is the velocity of projection of the ball

Answer 1

It takes 2s to reach maximum height.
Hence,
uSinx/g = 2
uSinx = 2g ———(1)

After 1s angle is 45°. At this point vertical component of velocity is equal to horizontal component of initial velocity.
Along vertical
v = u + at
vSin45 = uSinx - g
uCosx = uSinx - g
uCosx = 2g - g >>>>>>[From (1)]
uCosx = g ————(2)

Square and add equations (1) and (2)
u² = 4g² + g²
u = g√5
u = 9.8√5 m/s

Velocity of projection of ball is 9.8√5 m/s (or 10√5 m/s if g = 10m/s²)