Question

A ball is thrown from a height h metre with an initial downward velocity Vo it hits the ground and loses half of its kinetic energy and bounces back to same height what is the value of Vo

Answer 1

height h=10m

Initial velocity = V°

let m be the mass of ball.

Ball is projected downward with velocity V° from height h.

Total energy =mgh +1/2mv°²

Total Energy collision = 50/100x mgh + 1/2 mv°²

the ball rises back to same height after collision hence

V°=√2gh

   =√2x9.8x 10

   =14 m/s

therefore initial  velocity = 14m/s 

Answer 2

$\huge \underline \bold \color{red} \mathfrak{Answer }$

Height = h

Initial velocity = v

Let the mass of ball be m.

$Initial \: kinetic \: energy = \frac{1}{2} m {v}^{2}$

Final kinetic energy is half of the initial kinetic energy.

$\therefore \: Final \: kinetic \: energy = \frac{1}{2} m {v}^{2} \times \frac{1}{2} = \frac{1}{4} m {v}^{2}$

Now applying Energy Conservation Law

Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy

$\frac{1}{2} m {v}^{2} + mgh = \frac{1}{4} m {v}^{2} + 0$

$\frac{1}{2} {v}^{2} + gh = \frac{1}{4} {v}^{2}$

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

${v}^{2} = 2gh$

$\color{green} \bold{v = \sqrt{2gh} }$