Answer:
Option A
[tex]\frac{V}{\sqrt{2} }[/tex] ,[tex]\frac{E}{2}[/tex]
Explanation:
When the ball its thrown up, at half way of its flight it means half of its vertical height which is [tex]\frac{h}{2}[/tex].
potential energy = mgh
since it moved half way of height
P.E = [tex]\frac{mgh}{2}[/tex]
This means for the body to have gained half of its P.E, it will loose half of its kinetic energy.
Final kinetic energy([tex]E_{1}[/tex]) = E/2
kinetic energy = [tex]\frac{1}{2}mv^{2}[/tex]
let the final velocity at halfway flight be v1
[tex]E_{1}[/tex] = E/2
[tex]\frac{1}{2}mv_{1}^{2}[/tex] =[tex]\frac{\frac{1}{2}mv^{2}}{2}[/tex]
cross multiply we have
[tex]mv_{1}^{2}[/tex] =[tex]\frac{1}{2}mv^{2}[/tex]
cancel m from both sides
[tex]v_{1}^{2}[/tex] = [tex]\frac{1}{2}v^{2}[/tex]
take the square root of both sides,
[tex]v_{1} =\sqrt{\frac{v^{2} }{2} }[/tex]
[tex]v_{1} =\frac{v}{\sqrt{2} }[/tex]
Thus our final velocities will be E/2 and [tex] \frac{v}{\sqrt{2} }[/tex]
