A body is thrown horizontally with a velocity u from the top of a tower. The displacement of the stone when the horizontal and vertical velocity are equal is

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

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

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\sqrt{5} (\frac{u^{2}}{2 g})

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

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Solution

The correct option is B $\sqrt{5} (\frac{u^{2}}{2 g})$
We have,
$\begin{matrix} u^{2} = 2 g h h = \frac{u^{2}}{2 g} T = \sqrt{\frac{2 h}{g}} s = 4 \sqrt{\frac{2 h}{g}} = 4 \sqrt{\frac{2 h}{g} \times \frac{u^{2}}{2 g}} s = \frac{u^{2}}{2 g} D = \sqrt{h^{2} + s^{2}} = \frac{\sqrt{5 u^{2}}}{2 g} \end{matrix}$
Then,
Option $C$ is correct answer.

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Similar questions

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 $\frac{u^{2}}{2 g}$ . [ $g$ is the acceleration due to gravity]

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

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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A body is thrown horizontally with a velocity u from the top of a tower. The displacement of the stone when the horizontal and vertical velocity are equal is

The correct option is B √5(u22g)We have,u2=2ghh=u22gT=√2hgs=4√2hg=4√2hg×u22gs=u22gD=√h2+s2=√5u22gThen,Option C is correct answer.