A ball is thrown vertically upwards with a velocity $u$ . What is the maximum height to which it will rise before falling back?

u^{2} / 2 g

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g / u

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2 g / u

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u / 2 g

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Solution

The correct option is A $u^{2} / 2 g$
Answer is A.
If we know the initial velocity, from the equation, $v^{2} - u^{2} = 2 a s$ with s = h and v = 0 at the top of the path, initial velocity is given as 'u' and a = -g then:
h = $\frac{u^{2}}{2 g}$ .
Therefore, the maximum height reached = $\frac{u^{2}}{2 g}$ .

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If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ :

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