A ball of mass 0.2 ,kg is thrown vertically d | vedclass

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(B) Given: Mass of the ball $m = 0.2 \,kg$, height $h = 10 \,m$, acceleration due to gravity $g = 10 \,ms^{-2}$.Total energy just before collision at

Vedclass · Accepted Answer

$14 \,ms^{-1}$

Vedclass · Answer

(B) Given: Mass of the ball $m = 0.2 \,kg$, height $h = 10 \,m$, acceleration due to gravity $g = 10 \,ms^{-2}$.Total energy just before collision at the floor is $E_i = \frac{1}{2}mu^2 + mgh$.After collision, the ball loses $50 \%$ of its energy, so the remaining energy is $E_f = \frac{E_i}{2}$.The ball rises back to the same height $h$, meaning its potential energy at the peak is $mgh$.Equating the remaining energy to the potential energy at the peak: $\frac{1}{2}(\frac{1}{2}mu^2 + mgh) = mgh$.$\frac{1}{2}mu^2 + mgh = 2mgh$.$\frac{1}{2}mu^2 = mgh$.$u^2 = 2gh$.$u = \sqrt{2 	imes 10 	imes 10} = \sqrt{200} \approx 14.14 \,ms^{-1}$.Rounding to the nearest provided option, the initial velocity is $14 \,ms^{-1}$.

$A$ ball of mass $0.2 \,kg$ is thrown vertically down from a height of $10 \,m$. It collides with the floor and loses $50 \%$ of its energy and then rises back to the same height. The value of its initial velocity is

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