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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 :

Answer:

[tex]v^{2}_{2}=v^{2}_{1}+\frac{v^{2}Sin^{2}\theta }{g^{2}}[/tex]

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

[tex]v'^{2}=u^{2}+2as[/tex]

[tex]\left ( \frac{vSin\theta }{g} \right )^{2}=0^{2}+2gh[/tex]

[tex]h = \frac{v^{2}Sin^{2}\theta }{2g^{3}}[/tex]      . ... (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

[tex]v^{2}=u^{2}+2as[/tex]

[tex]v^{2}_{2}=v^{2}_{1}+2gh[/tex]

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

[tex]v^{2}_{2}=v^{2}_{1}+2g\frac{v^{2}Sin^{2}\theta }{2g^{3}}[/tex]

[tex]v^{2}_{2}=v^{2}_{1}+\frac{v^{2}Sin^{2}\theta }{g^{2}}[/tex]

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