An object of unknown mass is initially at rest and dropped from a height h. it reaches the ground with a velocity v sin θ / g 1. the same object is then raised again to the same height h but this time is thrown downward with velocity υ1 it now reaches the ground with a new velocity υ2. how is v2 related to v1?

Explanation:

Case I:

initial velocity, u = 0 m/s

Final velocity, v' = v Sinθ /g

Height = h

acceleration = g

Use third equation of motion, we get

     . ... (1)

Case II:

initial velocity, u = v1

Final velocity, v = v2

height = h

acceleration due to gravity = g

Use third equation of motion, we get

Substitute the value of h from equation (1) ,we get

measuring things 1 litre of water