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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\frac{u}{2 g}

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2 u^{2} g

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None of these

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Solution

The correct option is A $\frac{u^{2}}{2 g}$
From $v^{2} = u^{2} = 2 a s$
$∴ 0 = u^{2} = 2 g h$
$\Rightarrow h = \frac{u^{2}}{2 g}$

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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 $\frac{u^{2}}{2 g}$ :

If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ . [ $g$ is the acceleration due to gravity]

Q. If a body is thrown up with an initial velocity u and covers a height of h, then

h = \frac{u^{2}}{2 g}

If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to

Q. Assertion :The maximum height reached by an object projected vertically up is directly proportional to the initial velocity u. Reason: The maximum height reached by an object thrown up with an initial velocity u is given by

h = \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

The correct option is A u22gFrom v2=u2=2as∴0=u2=2gh⇒h=u22g

The correct option is A $\frac{u^{2}}{2 g}$
From $v^{2} = u^{2} = 2 a s$
$∴ 0 = u^{2} = 2 g h$
$\Rightarrow h = \frac{u^{2}}{2 g}$