Question

a particle is projected from the ground at angle theta with the horizontal .after 1s it is moving at angle 45degree with the horizontal and after 2s it is moving horizontally .what is the velocity of projection of the ball

Answer 1

The average velocity in a uniformly accelerated motion is given by Displacement/Time.

The displacement will be: [(v^2)*sin(2phi)/2g]i + [(v^2)*{sin(phi)}^2/2g]j

Magnitude of displacement: (v^2/2g)*[4{sin(phi)cos(phi)}^2 + {sin(phi)}^4]^0.5

Time: v*sin(phi)/g

Average velocity: (v/2)*{4(cos(phi))^2 + (sin(phi))^2}, after applying given formula.

Simplifying: (v/2)*{1 + 3(cos(phi))^2}^0.5