Question #59798

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Answer 1 · 2016-06-08T17:17:44

$E_2 = 0.27$ [J]

Just before the strike

$m g h_1 = E_1 = 0.75$ [J] then

$mg = 0.75/h_1 = 0.75/0.8$

After the strike the ball needs kinetic energy $E_2$ to attain $0.45$ [m]

so we have

$h_2 mg = E_2$ or

$E_2 = 0.45xx 0.75/0.8 = 0.27$ [J]