Question

A ball is thrown vertically upwards with a velocity v and an initial kinetic energy Ek. When half way to the top of its flight, it has a velocity and kinetic energy respectively of (A) v/square root 2, E/2 (B) v/4, E/2 or (C) v/2, E/square root 2 ?

Answer 1

Answer:

Option A

$\frac{V}{\sqrt{2} }$ ,$\frac{E}{2}$

Explanation:

When the ball its thrown up, at half way of its flight it means half of its vertical height which is $\frac{h}{2}$.  

potential energy = mgh

since it moved half way of height

P.E = $\frac{mgh}{2}$

This means for the body to have gained half of its P.E, it will loose half of its kinetic energy.

Final kinetic energy($E_{1}$) = E/2

kinetic energy = $\frac{1}{2}mv^{2}$

let the final velocity at halfway flight be v1

$E_{1}$ = E/2

$\frac{1}{2}mv_{1}^{2}$ =$\frac{\frac{1}{2}mv^{2}}{2}$

cross multiply we have

$mv_{1}^{2}$ =$\frac{1}{2}mv^{2}$

cancel m from both sides

$v_{1}^{2}$ = $\frac{1}{2}v^{2}$

take the square root of both sides,

$v_{1} =\sqrt{\frac{v^{2} }{2} }$

$v_{1} =\frac{v}{\sqrt{2} }$

Thus our final velocities will be E/2 and $\frac{v}{\sqrt{2} }$