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

u²/2g

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u/2g

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Solution

The correct option is A
u²/2g

Using equation $v^{2}$ = $u^{2} + 2 a s$ ,
Here, $a = - g$ , $∵$ acceleration due to gravity is acting opposite to the direction of motion.
$s = h$

At highest point $v = 0$ ,

$0 = u^{2} - 2 g h$

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

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Similar questions

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}

Q. Assertion :Maximum possible height attained by a projectile is

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

, where u is initial velocity Reason: To attain maximum height, a particle is thrown vertically upwards at

θ = 90^{o}

to the ground.

H = \frac{u^{2}}{2 g} s i n^{2} θ \Rightarrow H_{m a x} = \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 u22g:

The correct option is A u2/2g Using equation v2 = u2+2as, Here, a=−g, ∵ acceleration due to gravity is acting opposite to the direction of motion. s=h At highest point v=0, 0=u2−2gh h=u22g

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

The correct option is A
u²/2g

Using equation $v^{2}$ = $u^{2} + 2 a s$ ,
Here, $a = - g$ , $∵$ acceleration due to gravity is acting opposite to the direction of motion.
$s = h$

At highest point $v = 0$ ,

$0 = u^{2} - 2 g h$

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