JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Kinematics

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h . Find the ratio of the times in which it is at height

\frac{h}{3}

while going up and coming down respectively.

Answer: (B)

Solution

From maximum height

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

, we get

u = 2 g h

. Using

s = u t - \frac{1}{2} g t^{2}

with

s = \frac{h}{3}

\frac{h}{3} = 2 g h t - \frac{1}{2} g t^{2}

. Rearranging:

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

. The two roots

t_{1}

(going up) and

t_{2}

(coming down) satisfy:

t_{1} + t_{2} = \frac{2 2 g h}{g}

and

t_{1} t_{2} = \frac{2 h}{3 g}

. Using the quadratic formula,

t_{1} = \frac{2 g h - 2 g h - \frac{2 g h}{3}}{g} = \frac{2 g h - \frac{4 g h}{3}}{g}

and

t_{2} = \frac{2 g h + \frac{4 g h}{3}}{g}

. The ratio

\frac{t _{1}}{t _{2}} = \frac{2 g h - \frac{4 g h}{3}}{2 g h + \frac{4 g h}{3}} = \frac{3 - 2}{3 + 2}

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option