JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Laws of Motion

A small ball of mass

m

is thrown upward with velocity

u

from the ground. The ball experiences a resistive force

mk v^{2}

where

v

is its speed. The maximum height attained by the ball is:

Answer: (D)

Solution

On the way up the ball feels its weight

m g

and the resistive force

mk v^{2}

, both acting downward, so

v \frac{d v}{d x} = - (g + k v^{2})

. Separating variables and integrating from launch (

v = u

x = 0

) to the top (

v = 0

x = h

\int_{u}^{0} \frac{v d v}{g + k v ^{2}} = - \int_{0}^{h} d x

, which gives

h = \frac{1}{2 k} ln (\frac{g + k u ^{2}}{g}) = \frac{1}{2 k} ln (1 + \frac{k u ^{2}}{g})

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option