Question

a particle projected from ground moves at angle 45°with horizontal one second after projection and speed is minimum, two seconds after projection. the angle of projection of particle is​

Answer 1

Answer:

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