A ball is thrown vertically downwards from a height of $$20$$ m with an initial velocity ν$$0$$. It collides with the ground, loses $$50$$ percent of its energy in collision and rebounds to the same height. The initial velocity ν$$0$$ is (Take$$ g $$=$$ $$10$$ $$ms-2$$ $$)

Question

A ball is thrown vertically downwards from a height of $$20$$ m with an initial velocity ν$$0$$. It collides with the ground, loses $$50$$ percent of its energy in collision and rebounds to the same height. The initial velocity ν$$0$$ is (Take$$ g $$=$$ $$10$$ $$ms-2$$ $$)

Infinity Learn · Accepted Answer

The situation is shown in the figureLet ν be the velocity of the ball with which it collides with ground. Then according to the law of conservation of energy,Gain in kinetic energy $$=$$ loss in potential energy  i.e. $$12mv2-12mv02=$$$$mgh$$ (where$$ m is the mass of the $$ball)  or   $$v2-v02=2gh$$              . . . .(i)Now$$, when the ball collides with the ground, $$50$$% of its energy is lost and it rebounds to the same height h.∴  $$5010012mv2=mgh14v2=gh$$   or   $$v2=4gh$$ Substituting this value of $$v2$$ in eqn. $$(i), we get $$4gh-v02=2gh$$ or   $$v02=4gh-2gh=2gh$$ or $$v0=2gh$$ Here, $$g=10$$ $$ms-2and$$ $$h=20$$ m∴  $$v0=210$$ $$ms-2(20$$ $$m)=20$$ $$ms-1$$

A ball is thrown vertically downwards from a height of 20 m with an initial velocity $ν_{0}$ . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $ν_{0}$ is (Take g = $10 {ms}^{- 2}$ )

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A ball is thrown vertically downwards from a height of 20 m with an initial velocity ν0. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity ν0 is (Take g = 10 ms-2 )

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A ball is thrown vertically downwards from a height of 20 m with an initial velocity $ν_{0}$ . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $ν_{0}$ is (Take g = $10 {ms}^{- 2}$ )