Question

A particle projected from ground moves at an angle 45 degrees at horizontal one second after projection and speed is minimum two seconds after the projection. The angle of projection of particle is

Answer 1

hey mate !!!

It depends on what value you use for g. I'll use g=-10m/s² (with the convetion upwards is positive, negative is downwards). 

But you might be expected to use -9.8m/s² or -9.81m/s²; of course, the method is the same. 

Also we must assume air resistance is negligible. 
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The speed is minimum at the highest point. This is because at the highest point, the vertical component of velocity = 0 and the horizontal component of velocity is constant during the flight. 

If the initial vertical component of velocity is uy, during the first 2s the vertical component of velocity drops from uy to 0. 

Using the standard equation 'v = u+at' for the vertical motion: 
0 = uy + at 
0 = uy + (-10)*2 
uy = 20m/s 
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After 1s: 
vy = uy + at 
vy = 20 + (-10)*1 
vy = 10m/s 

After 1s, the direction is 45° to the horizontal so the x and y components of velocity (vx, vy) are equal. So vx = vy = 10m/s. 
vx is constant throughout the flight. 

The initial angle of launch, θ, is given by: 
tanθ = uy/vx = 20/10 = 2 
θ = tan⁻¹(2) = 63°