Question 151: Laws of Motion

Question

A ball is thrown upward with an initial velocity
V
0
from the surface of the earth. The motion
of the ball is affected by a drag force equal to
m$$\gamma $$u
2
(where m is mass of the ball, u is its
instantaneous velocity and $$\gamma $$ is a constant).
Time taken by the ball to rise to its zenith is :

Accepted Answer

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
$a_{	ext{net}} = - (g + \gamma v^2) = \frac{dv}{dt}$, 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 :
$$\frac{1}{\gamma} \int\limits_{v_0}^{0} \frac{1}{\left(\frac{g}{\gamma}+v^2ight)} dv = - \int\limits_{0}^{t} dt$$
Recognizing the integral as the standard form $\frac{1}{x^2 + a^2} = \frac{1}{a} 	an^{-1}\left(\frac{x}{a}ight)$, we write the integral in terms of the arctangent function.
This gives us :
$$\frac{1}{\gamma} \left(\frac{1}{\sqrt{\frac{g}{\gamma}}}ight) \left[	an^{-1}\left(\frac{v}{\sqrt{\frac{g}{\gamma}}}ight)ight]_{v_0}^{0} = -t$$
Evaluating the integral at the bounds gives us :
$$\frac{1}{\sqrt{\gamma g}} 	an^{-1}\left(\frac{\sqrt{\gamma} v_0}{\sqrt{g}}ight) = t$$
Therefore, the time taken by the ball to rise to its zenith, considering the drag force, is given by
$$t = \frac{1}{\sqrt{\gamma g}} 	an^{-1}\left(\sqrt{\frac{\gamma}{g}} V_0ight)$$

Answer

$${1 \over {\sqrt {\gamma g} }}{	an ^{ - 1}}\left( {\sqrt {{\gamma  \over g}} {V_0}} ight)$$

Answer

$${1 \over {\sqrt {\gamma g} }}{ln}\left( 1+ {\sqrt {{\gamma  \over g}} {V_0}} \right)$$

Answer

$${1 \over {\sqrt {\gamma g} }}{\sin ^{ - 1}}\left( {\sqrt {{\gamma  \over g}} {V_0}} \right)$$

Answer

$${1 \over {\sqrt {2\gamma g} }}{	an ^{ - 1}}\left( {\sqrt {{2\gamma  \over g}} {V_0}} ight)$$

Laws of Motion

A ball is thrown upward with an initial velocity V 0 from the surface of the earth. The motion of the ball is affected by a drag force equal to m$\gamma u 2 (where m is mass of the ball, u is its instantaneous velocity and \gamma $ is a constant). Time taken by the ball to rise to its zenith is :

Choose the correct answer: