QJEEM19Ph4921: JEE Main 2019 Physics - Class 11 JEEM19Ph4921 Solution: A ball is thrown upward with an initial velocity V 0 from the…

Given, drag force, $F=m \gamma v^2$ ......(i)

As we know, general equation of force

$=m a$ .........(ii)

Comparing Eqs. (i) and (ii), we get

$a=\gamma v^2$

The net retardation of the ball when thrown vertically upward is therefore

$a_{\text{net}} = - (g + \gamma v^2) = \frac{dv}{dt}$ , where $g$ is the acceleration due to gravity.

Rearranging terms gives us :

$\frac{dv}{g + \gamma v^2} = - dt$

We now need to integrate both sides of this equation.

When the ball is thrown upward with velocity $v_0$ and reaches its zenith ( "zenith" refers to the highest point that the ball reaches in its upward trajectory.), the velocity is $0$ . The time to reach the zenith is $t$ .

So the integral equation is :

$\int\limits_{v_0}^{0} \frac{dv}{\gamma v^2 + g} = - \int\limits_{0}^{t} dt$

Separating the constants from the integral :

$\frac{1}{\gamma} \int\limits_{v_0}^{0} \frac{1}{\left(\frac{g}{\gamma}+v^2\right)} dv = - \int\limits_{0}^{t} dt$

Recognizing the integral as the standard form $\frac{1}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$ , we write the integral in terms of the arctangent function.

This gives us :

$\frac{1}{\gamma} \left(\frac{1}{\sqrt{\frac{g}{\gamma}}}\right) \left[\tan^{-1}\left(\frac{v}{\sqrt{\frac{g}{\gamma}}}\right)\right]_{v_0}^{0} = -t$

Evaluating the integral at the bounds gives us :

$\frac{1}{\sqrt{\gamma g}} \tan^{-1}\left(\frac{\sqrt{\gamma} v_0}{\sqrt{g}}\right) = t$

Therefore, the time taken by the ball to rise to its zenith, considering the drag force, is given by

$t = \frac{1}{\sqrt{\gamma g}} \tan^{-1}\left(\sqrt{\frac{\gamma}{g}} V_0\right)$