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 \[m\gamma {v^2}\] (where \[m\] is the mass of the ball, \[v\] is its instantaneous velocity and \[\gamma \] is a constant). Time taken by the ball to rise to its zenith is:
(A) \[\dfrac{1}{{\sqrt {\gamma g} }}{\sin ^{ - 1}}\left( {\sqrt {\dfrac{\gamma }{g}} } \right){V_0}\]
(B) \[\dfrac{1}{{\sqrt {\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {\dfrac{\gamma }{g}} } \right){V_0}\]
(C) \[\dfrac{1}{{\sqrt {2\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {\dfrac{{2\gamma }}{g}} } \right){V_0}\]
(D) \[\dfrac{1}{{\sqrt {\gamma g} }}\ln \left( {1 + \sqrt {\dfrac{\gamma }{g}} } \right){V_0}\]

Answer 1

Hint:First of all, we will find the net force acting on the ball, and from there we will simplify. After that we will integrate both the sides taking the respective limits followed by manipulation to find an expression for time taken.

Complete step by step answer:
In the given question, we are supplied with following data:
The velocity with which the ball is thrown upward is \[{V_0}\] and it can be termed as the initial velocity of the ball.
The drag force which affects the motion of the ball is represented as \[m\gamma {v^2}\] .
We are asked to find the time taken by the ball to rise to the highest position.
To begin with, we will find the net force on the body, which can be done with the following manipulation:
But before we continue, we must know that the weight of the ball and the drag force is acting in a downward direction. The net force on the body is acting in the upward direction.
So, by balancing the forces we can write mathematically as:
$
ma = - mg - m\gamma {v^2} \\
a = - \left( {g + \gamma {v^2}} \right) \\
$
Again, we can write:
Acceleration is the rate of change of velocity, which we can write mathematically as:
\[a = \dfrac{{dv}}{{dt}}\]
So, using this in the above equation of acceleration, we get:
$a = - \left( {g + \gamma {v^2}} \right) \\
\Rightarrow\dfrac{{dv}}{{dt}} = - \left( {g + \gamma {v^2}} \right) \\
\Rightarrow\dfrac{{dv}}{{dt}} = - g\left( {1 + \dfrac{{\gamma {v^2}}}{g}} \right) \\$
\[ - gdt = \dfrac{{dv}}{{1 + \dfrac{{\gamma {v^2}}}{g}}}\] …… (1)
Now, we will integrate the equation (1), where time has an upper limit of \[t\] and lower limit of \[0\]. Again, the upper limit of velocity is \[0\] and the lower limit of velocity is \[{V_0}\] .
So, the equation (1) becomes:
\[\int_0^t { - gdt} = \int_{{V_0}}^0 {\dfrac{{dv}}{{1 + \dfrac{{\gamma {v^2}}}{g}}}} \] …… (2)
We will apply the integration formula which is given below:
\[\int_0^x {\dfrac{1}{{1 + {x^2}}}} = {\tan ^{ - 1}}x\]
Now, we will manipulate the equation (2) and we get:
$
\int_0^t { - gdt} = \int_{{V_0}}^0 {\dfrac{{dv}}{{1 + \dfrac{{\gamma {v^2}}}{g}}}} \\
\Rightarrow - g \times \int_0^t {dt} = \int_{{V_0}}^0 {\dfrac{{dv}}{{1 + {{\left( {\dfrac{{\sqrt \gamma v}}{{\sqrt g }}} \right)}^2}}}} \\
\Rightarrow - g\left[ t \right]_0^t = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\sqrt \gamma v}}{{\sqrt g }}} \right)} \right]_{{V_0}}^0 \\
\Rightarrow - gt = - \sqrt {\dfrac{g}{\gamma }} {\tan ^{ - 1}}\sqrt {\dfrac{\gamma }{g}} {V_0} \\$
Again, manipulating we get:
\[t = \sqrt {\dfrac{1}{{g\gamma }}} {\tan ^{ - 1}}\sqrt {\dfrac{\gamma }{g}} {V_0}\]

Hence, the required by the ball to rise to its zenith is \[\sqrt {\dfrac{1}{{g\gamma }}} {\tan ^{ - 1}}\sqrt {\dfrac{\gamma }{g}} {V_0}\]. Thus the correct option is B.

Note:While solving this problem, you should have a firm knowledge of integration. Most of the students tend to make mistakes while choosing the upper and the lower limits. The initial velocity is the lower limit and the velocity of the ball at the zenith position will become zero, which is the upper limit.