Question

One second after projection, a stone moves at an
ole of 45° with the horizontal. Two seconds after
projection, it moves horizontally. Its angle of
projection is (in degrees) (g = 10 m/s2)
(1) 60°
ature
(2) tan"(4)
(3) tan (3)
(4) tan (2)​

Answer 1

X = u cos θ t                 and   Vx = u cos θ

y = u sin θ t - 1/2 g t²     abd   Vy = u sin θ - g t

The horizontal component of speed Vx is always constant.  So the speed is minimum when the vertical component Vy is 0.

=> u sinθ = 2 g 

45° = direction of velocity of the project at  time t = 1 sec.

Then  Tan 45° = 1 = Vy / Vx = (u sinθ - g*1)/(u cosθ)

=> u cosθ = g

=>  tan θ = 2    =>  θ = Tan⁻¹ 2