A ball is thrown vertically upward. It has speed of $10 m / s$ when it is reached one half of its maximum height. How high does the ball rise? (Taking $g = 10 m / s^{2}$

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Solution

From third equation of motion , $v^{2} = u^{2} - 2 g h$
$0 = u^{2} - 2 g (h / 2)$
$h = \frac{u^{2}}{g} = \frac{10^{2}}{10} = 10 m$
Thus, ball rises to 10m

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A ball is thrown vertically upward. It has speed of 10m/s when it is reached one half of its maximum height. How high does the ball rise? (Taking g=10m/s2

From third equation of motion ,v2=u2−2gh0=u2−2g(h/2)h=u2g=10210=10mThus, ball rises to 10m

A ball is thrown vertically upward. It has speed of $10 m / s$ when it is reached one half of its maximum height. How high does the ball rise? (Taking $g = 10 m / s^{2}$

From third equation of motion , $v^{2} = u^{2} - 2 g h$
$0 = u^{2} - 2 g (h / 2)$
$h = \frac{u^{2}}{g} = \frac{10^{2}}{10} = 10 m$
Thus, ball rises to 10m