JEE Main 2019 — Laws Of Motion Question with Solution

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

As we know, general equation of force

$$=ma$$ .........(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 :

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

Separating the constants from the integral :

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

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

$$t=\frac{1}{\sqrt{\gamma g}}{\tan }^{-1}\left(\sqrt{\frac{\gamma }{g}}{V}_0\right)$$