Question

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?

Answer 1

`v^(2)_(2)=v^(2)_(1)+(v^(2)Sin^(2)\theta )/(g^(2))`

Step-by-step 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

`v'^(2)=u^(2)+2as`

`\left ( (vSin\theta )/(g) \right )^(2)=0^(2)+2gh`

`h = (v^(2)Sin^(2)\theta )/(2g^(3))` . ... (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

`v^(2)=u^(2)+2as`

`v^(2)_(2)=v^(2)_(1)+2gh`

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

`v^(2)_(2)=v^(2)_(1)+2g(v^(2)Sin^(2)\theta )/(2g^(3))`

`v^(2)_(2)=v^(2)_(1)+(v^(2)Sin^(2)\theta )/(g^(2))`