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

Answer:

$v^{2}_{2}=v^{2}_{1}+\frac{v^{2}Sin^{2}\theta }{g^{2}}$

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

$v'^{2}=u^{2}+2as$

$\left ( \frac{vSin\theta }{g} \right )^{2}=0^{2}+2gh$

$h = \frac{v^{2}Sin^{2}\theta }{2g^{3}}$      . ... (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

$v^{2}_{2}=v^{2}_{1}+2g\frac{v^{2}Sin^{2}\theta }{2g^{3}}$

$v^{2}_{2}=v^{2}_{1}+\frac{v^{2}Sin^{2}\theta }{g^{2}}$