JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Kinematics

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

\frac{h}{3}

in both the directions.

Answer: (B)

Solution

The ball leaves with

u = 2 g h

because it rises exactly to the height

h

. The height

h /3

is crossed when

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

, a quadratic whose two roots are the times going up and coming down:

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

Their ratio is

\frac{t _{1}}{t _{2}} = \frac{2 - \frac{2}{3}}{2 + \frac{2}{3}} = \frac{6 - 2}{6 + 2} = 5 - 26

and

\frac{3 - 2}{3 + 2} = (3 - 2)^{2} = 5 - 26

as well, so the two times are in the ratio

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

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option