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(B) Let the initial velocity be $u$ and the maximum height be $H$. The maximum height is given by $H = \frac{u^2}{2g}$.At height $h = \frac{H}{2}$,the

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$10$

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(B) Let the initial velocity be $u$ and the maximum height be $H$. The maximum height is given by $H = \frac{u^2}{2g}$.At height $h = \frac{H}{2}$,the velocity $v = 10 \; m/s$.Using the equation of motion $v^2 = u^2 - 2gh$:$(10)^2 = u^2 - 2g \left( \frac{H}{2} ight)$$100 = u^2 - gH$Substitute $H = \frac{u^2}{2g}$ into the equation:$100 = u^2 - g \left( \frac{u^2}{2g} ight)$$100 = u^2 - \frac{u^2}{2} = \frac{u^2}{2}$$u^2 = 200 \; m^2/s^2$.Now,calculate the maximum height $H$:$H = \frac{u^2}{2g} = \frac{200}{2 	imes 10} = \frac{200}{20} = 10 \; m$.

$A$ ball is thrown vertically upward. It has a speed of $10 \; m/s$ when it has reached one-half of its maximum height. How high does the ball rise? (Take $g = 10 \; m/s^2$)

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