A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching h \over 3 in both the directions.

\sqrt 3 - \sqrt 2 \over \sqrt 3 + \sqrt 2

Motion in A Straight Line PYQ — JEE Main 2021 · 27 Jul · Shift 1 · Q62 with Solution

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Correct answer: C

Maximum height relation

If the ball is thrown upward with initial speed $u$ , then its maximum height is

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

So,

u^2 = 2gh

Equation of motion for height $\dfrac{h}{3}$

At time $t$ , height is

y = ut - \frac{1}{2}gt^2

We want the times when the ball is at height $\dfrac{h}{3}$ :

ut - \frac{1}{2}gt^2 = \frac{h}{3}

Using $h = \dfrac{u^2}{2g}$ ,

ut - \frac{1}{2}gt^2 = \frac{1}{3}\cdot \frac{u^2}{2g} = \frac{u^2}{6g}

Multiply by $2$ :

2ut - gt^2 = \frac{u^2}{3g}

Multiply throughout by $g$ :

2ugt - g^2 t^2 = \frac{u^2}{3}

Rearrange:

g^2 t^2 - 2ugt + \frac{u^2}{3} = 0

Solve quadratic in $t$

Using quadratic formula,

t = \frac{2ug \pm \sqrt{(2ug)^2 - 4g^2\cdot \frac{u^2}{3}}}{2g^2}

t = \frac{2ug \pm \sqrt{4u^2g^2 - \frac{4u^2g^2}{3}}}{2g^2}

t = \frac{2ug \pm 2ug\sqrt{\frac{2}{3}}}{2g^2}

t = \frac{u}{g}\left(1 \pm \sqrt{\frac{2}{3}}\right)

Thus the two times are

t_1 = \frac{u}{g}\left(1 - \sqrt{\frac{2}{3}}\right), \qquad t_2 = \frac{u}{g}\left(1 + \sqrt{\frac{2}{3}}\right)

where $t_1$ is during upward motion and $t_2$ during downward motion.

Required ratio

The ratio of the two times is

\frac{t_1}{t_2} = \frac{1 - \sqrt{2/3}}{1 + \sqrt{2/3}}

Multiply numerator and denominator by $\sqrt{3}$ :

\frac{t_1}{t_2} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}

Match with options

This matches Option C.

\boxed{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}

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View written solutionFree

Correct answer: C

Maximum height relation

If the ball is thrown upward with initial speed $u$ , then its maximum height is

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

So,

u^2 = 2gh

Equation of motion for height $\dfrac{h}{3}$

At time $t$ , height is

y = ut - \frac{1}{2}gt^2

We want the times when the ball is at height $\dfrac{h}{3}$ :

ut - \frac{1}{2}gt^2 = \frac{h}{3}

Using $h = \dfrac{u^2}{2g}$ ,

ut - \frac{1}{2}gt^2 = \frac{1}{3}\cdot \frac{u^2}{2g} = \frac{u^2}{6g}

Multiply by $2$ :

2ut - gt^2 = \frac{u^2}{3g}

Multiply throughout by $g$ :

2ugt - g^2 t^2 = \frac{u^2}{3}

Rearrange:

g^2 t^2 - 2ugt + \frac{u^2}{3} = 0

Solve quadratic in $t$

Using quadratic formula,

t = \frac{2ug \pm \sqrt{(2ug)^2 - 4g^2\cdot \frac{u^2}{3}}}{2g^2}

t = \frac{2ug \pm \sqrt{4u^2g^2 - \frac{4u^2g^2}{3}}}{2g^2}

t = \frac{2ug \pm 2ug\sqrt{\frac{2}{3}}}{2g^2}

t = \frac{u}{g}\left(1 \pm \sqrt{\frac{2}{3}}\right)

Thus the two times are

t_1 = \frac{u}{g}\left(1 - \sqrt{\frac{2}{3}}\right), \qquad t_2 = \frac{u}{g}\left(1 + \sqrt{\frac{2}{3}}\right)

where $t_1$ is during upward motion and $t_2$ during downward motion.

Required ratio

The ratio of the two times is

\frac{t_1}{t_2} = \frac{1 - \sqrt{2/3}}{1 + \sqrt{2/3}}

Multiply numerator and denominator by $\sqrt{3}$ :

\frac{t_1}{t_2} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}

Match with options

This matches Option C.

\boxed{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}

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