A body is thrown vertically upwards. At half | vedclass

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(B) Let the maximum height attained be $H$. The initial velocity is $u$. At maximum height,the final velocity is $0$.Using the equation $v^2 = u^2 - 2

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$10$

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(B) Let the maximum height attained be $H$. The initial velocity is $u$. At maximum height,the final velocity is $0$.Using the equation $v^2 = u^2 - 2gh$,at maximum height $H$,we have $0 = u^2 - 2gH$,which gives $u^2 = 2gH$.At half the maximum height,$h = H/2$,the velocity is $v = 10 \ m/s$.Substituting these into the equation $v^2 = u^2 - 2gh$:$(10)^2 = u^2 - 2g(H/2)$$100 = u^2 - gH$Since $u^2 = 2gH$,we substitute this into the equation:$100 = 2gH - gH$$100 = gH$Given $g = 10 \ m/s^2$,we have $100 = 10 	imes H$.Therefore,$H = 10 \ m$.

$A$ body is thrown vertically upwards. At half of the maximum height,its velocity is $10 \ m/s$. What is the maximum height attained by the body? (Take $g = 10 \ m/s^2$)

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