Question

send Handwritten pic plz..​

Answer 1

Answer:

2h/5

Here is ur ans.......

Thnks:)

Answer 2

Solution:-

$Let \: us \: first \: take \: out \: the \: velocity \: just \\ \: before \: touching \: the \: groung \\ \\ Initil \: velocity \: of \: ball = 0 \\ acceleration = g \\ Distance \: it \:have \: to \: travel = h \\ Using \: third \: law \: of \: uniformly \: accelerated \\ motion \: we \: get \\ {v}^{2} = {0}^{2} + 2gh \\ v = \sqrt{2gh} \\ kinetic \: energy \: just \: before \: the \: ball \\ touches \: the \: ground = \frac{1}{2} m { (\sqrt{2gh} })^{2} = mgh \\ But \: 60percent \: of \: its \: energy \: is \: lost \\ Energy \: remains = \frac{40}{100} \times mgh = \frac{2mg}{5} \\ since \: mass \: is \: constant \: so \: the \: total \: energy \\ lost \: by \: ball \: is \: due \: to \: change \: in \: its \: kinetic \\ energy \\ let \: the \: velocity \: after \: lossing \: the \: energy \: be \: x \\ now \\ \frac{m {x}^{2} }{2} = \frac{2mgh}{5} \\ x = \sqrt{ \frac{4gh}{5} } \\ This \: will \: be \: the \: initial \: velocity \: in \: the \: case \: \\ when \: it \: bounds \: back \\ final \: velocity(v) = 0 \\ acceleration = g \\ using \: third \: law \: we \: get \\ {0}^{2} = \frac{4gh}{5} - 2gy \\ where \: y \: is \: the \: distance \: it \: move \: upward \\ \\ y = \frac{2h}{5} \\ Hence \: it \: moves \: back \: upto \: height \: of \: \frac{2h}{5}$

(hope it helps you)