Question

A ball is thrown upward with an initial velocity V₀ from the surface of the earth. The motion of the ball is affected by a drag force equal to mγv² (where m is mass of the ball, v is its instantaneous velocity and γ is a constant). Time taken by the ball to rise to its zenith is:
1. \(\frac{1}{{\sqrt {\gamma {\rm{g}}} }}{\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\sqrt {\frac{\gamma }{{\rm{g}}}} {{\rm{V}}_0}} \right)\)
2. \(\frac{1}{{\sqrt {\gamma {\rm{g}}} }}{\rm{si}}{{\rm{n}}^{ - 1}}\left( {\sqrt {\frac{\gamma }{g}} {{\rm{V}}_0}} \right)\)
3. \(\frac{1}{{\sqrt {\gamma {\rm{g}}} }}{\rm{ln}}\left( {1 + \sqrt {\frac{\gamma }{{\rm{g}}}} {{\rm{V}}_0}} \right)\)
4. \(\frac{1}{{\sqrt {2\gamma {\rm{g}}} }}{\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\sqrt {\frac{{2\gamma }}{{\rm{g}}}} {{\rm{V}}_0}} \right)\)

Answer 1

Correct Answer - Option 1 : \(\frac{1}{{\sqrt {\gamma {\rm{g}}} }}{\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\sqrt {\frac{\gamma }{{\rm{g}}}} {{\rm{V}}_0}} \right)\)

Concept:

Force:

The formula for force says force is equal to mass (m) multiplied by acceleration (a).

F = ma

The relationship between force and acceleration is given by

\(a = \frac{F}{m}\) 

The net force is the sum of n forces acting on an object.

Calculation:

Given,

Initial velocity of the ball, u = V₀

Drag force, F_d = mγv²

Velocity at a maximum height = 0

Where m is the Mass of the ball,

v is its instantaneous velocity

and γ is a constant

We know that acceleration is the change in velocity.

Force is given by,

F = ma

Thus, the net force on the ball:

F_net = mg + mγv²

F_net = m(g + γv²)

Acceleration is given by,

\(a = \frac{F}{m}\) 

\(a = \frac{{m\left( {g + \gamma {v^2}} \right)}}{m}\) 

a = -(g + γv²)

Thus, net acceleration is the change in the velocity,

\(\frac{{{\rm{dv}}}}{{{\rm{dt}}}} = {\rm{a}}\)

\(\frac{{dv}}{{dt}} = - \left( {{\rm{g}} + \gamma {{\rm{v}}^2}} \right)\) 

\(\frac{{dv}}{{ - \left( {{\rm{g}} + \gamma {{\rm{v}}^2}} \right)}} = dt\) 

\(\frac{{ - {\rm{dv}}}}{{{\rm{g}} + \gamma {{\rm{v}}^2}}} = dt\) 

Integrating both the sides,

\(\mathop \smallint \nolimits_{{{\rm{v}}_0}}^v \frac{{ - {\rm{dv}}}}{{{\rm{g}} + \gamma {{\rm{v}}^2}}} = \mathop \smallint \nolimits_0^{\rm{t}} {\rm{dt}}\) 

Let time t required to rise to its zenith (v = 0) so,

\( \Rightarrow \frac{1}{\gamma }\mathop \smallint \nolimits_{{{\rm{v}}_0}}^0 \frac{{ - {\rm{dv}}}}{{\frac{g}{\gamma } + {{\rm{v}}^2}}} = \mathop \smallint \nolimits_0^{\rm{t}} {\rm{dt}}\) 

We know that, according to integral formula,

\(\smallint \frac{1}{{{x^2} + {a^2}}}dx = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + C\) 

Thus,

\( \Rightarrow \frac{1}{\gamma }\mathop \smallint \nolimits_{{{\rm{v}}_0}}^0 \frac{{ - {\rm{dv}}}}{{{{\left( {\sqrt {\frac{{\rm{g}}}{{\rm{\gamma }}}} } \right)}^2} + {{\left( v \right)}^2}}} = \mathop \smallint \nolimits_0^{\rm{t}} {\rm{dt}}\) 

Here, x = \(\sqrt {\frac{{\rm{g}}}{\gamma }} \), a = v

\( \Rightarrow t = \frac{1}{\gamma }\left\{ {\left[ { - \frac{1}{{\sqrt {\frac{{\rm{g}}}{{\rm{\gamma }}}} }}{{\tan }^{ - 1}}\left( {\frac{v}{{\sqrt {\frac{{\rm{g}}}{{\rm{\gamma }}}} }}} \right)} \right]_{{V_0}}^0 + C} \right\}\) 

[for H_max, v = 0]

\(t = 0 - \frac{1}{\gamma }\left( { - \sqrt {\frac{{\rm{\gamma }}}{{\rm{g}}}} {{\tan }^{ - 1}}\left( {\frac{{\sqrt \gamma {V_0}}}{{\sqrt g }}} \right)} \right)\) 

\( \Rightarrow {\rm{t}} = \frac{1}{{\sqrt {\gamma {\rm{g}}} }}{\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt \gamma {V_0}}}{{\sqrt {\rm{g}} }}} \right)\) 

Therefore, the time taken by the ball to rise to its zenith is \(\frac{1}{{\sqrt {\gamma {\rm{g}}} }}{\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\sqrt {\frac{\gamma }{g}} {{\rm{V}}_0}} \right)\)