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Standard XII

Physics

3rd Equation of Motion

A body is thr...

Question

A body is thrown vertically upwards with an initial velocity $u$ . The expression for the maximum height attained by the body is

\frac{u}{2 g}

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

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

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

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Solution

The correct option is B $\frac{u^{2}}{2 g}$
From third equation of motion,
$v^{2} = u^{2} - 2 g S$
where $v = final velocity = 0 , u = initial velocity , S = distance$
$0 = u^{2} - 2 g S$
$u^{2} = 2 g S$
$S = \frac{u^{2}}{2 g}$

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Similar questions

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}

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}

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. 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 vertically upwards with an initial velocity u. The expression for the maximum height attained by the body is

The correct option is B u22gFrom third equation of motion, v2=u2−2gSwherev = final velocity = 0 , u = initial velocity , S = distance0=u2−2gSu2=2gSS=u22g

A body is thrown vertically upwards with an initial velocity $u$ . The expression for the maximum height attained by the body is

The correct option is B $\frac{u^{2}}{2 g}$
From third equation of motion,
$v^{2} = u^{2} - 2 g S$
where $v = final velocity = 0 , u = initial velocity , S = distance$
$0 = u^{2} - 2 g S$
$u^{2} = 2 g S$
$S = \frac{u^{2}}{2 g}$