A body is thrown up with an initial velocity u and it 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 g

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None of these

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Solution

The correct option is A $\frac{u^{2}}{2 g}$
At maximum height, velocity of body is zero i.e. $v = 0$
Initial velocity is $u$
Acceleration $a = - g$
Using $v^{2} - u^{2} = 2 a S$
$∴$ $0 - u^{2} = 2 (- g) h$
$⟹ h = \frac{u^{2}}{2 g}$

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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. Question 2
A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a)

\frac{u}{2}

(b)

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

(c)

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

(d)

\frac{u}{2 g}

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

The correct option is A u22gAt maximum height, velocity of body is zero i.e. v=0Initial velocity is uAcceleration a=−gUsing v2−u2=2aS∴ 0−u2=2(−g)h⟹ h=u22g

The correct option is A $\frac{u^{2}}{2 g}$
At maximum height, velocity of body is zero i.e. $v = 0$
Initial velocity is $u$
Acceleration $a = - g$
Using $v^{2} - u^{2} = 2 a S$
$∴$ $0 - u^{2} = 2 (- g) h$
$⟹ h = \frac{u^{2}}{2 g}$