© All Rights reserved @ LearnWithDash
Step-by-Step Solution
1. Identify the Forces Acting on the Ball
When the ball is thrown upwards with an initial velocity $V_0$, two forces act on it:
Its weight, $mg$, acting downward.
A drag force $F_\text{drag} = m \gamma v^2$, also acting opposite to the direction of motion (downward when the ball moves upward).
2. Write the Net Acceleration
As the ball rises, its velocity is directed upward and the drag force opposes this motion, also acting downward. The net downward acceleration can be written as:
$ a_\text{net} = - \bigl(g + \gamma v^2\bigr). $
Here, the negative sign indicates that the acceleration is downward while the velocity is upward (when the ball is rising).
3. Express Acceleration in Differential Form
By definition, acceleration is the time derivative of velocity:
$ a = \frac{dv}{dt}. $
So, combining with the net acceleration, we have:
$ \frac{dv}{dt} = - \bigl(g + \gamma v^2\bigr). $
Rewriting,
$ \frac{dv}{g + \gamma v^2} = -\,dt. $
4. Set Up the Integral with Appropriate Limits
To find the time taken to reach the highest point (the zenith), we integrate from the initial velocity $v_0$ (upwards) to the final velocity $0$ (at the top):
$ \int_{v_0}^{0} \frac{dv}{g + \gamma v^2} = - \int_{0}^{t} dt. $
On the right side, the time changes from $0$ to $t$, which is the time we seek.
5. Perform the Integration
First, factor out $ \gamma $ from the denominator on the left side:
$ \int_{v_0}^{0} \frac{dv}{g + \gamma v^2}
= \frac{1}{\gamma} \int_{v_0}^{0} \frac{dv}{\frac{g}{\gamma} + v^2}. $
Recognize that an integral of the form $ \int \frac{dx}{a^2 + x^2} $ is $ \frac{1}{a} \tan^{-1}\!\bigl(\frac{x}{a}\bigr) $. Here, $ a = \sqrt{\frac{g}{\gamma}} $.
Thus,
$ \frac{1}{\gamma} \int_{v_0}^{0} \frac{dv}{\frac{g}{\gamma} + v^2}
= \frac{1}{\gamma} \cdot \frac{1}{\sqrt{\frac{g}{\gamma}}}
\left[ \tan^{-1}\!\Bigl(\frac{v}{\sqrt{\frac{g}{\gamma}}}\Bigr) \right]_{v_0}^{0}. $
6. Evaluate the Integral at the Limits
Substitute the limits $v = 0$ and $v = v_0$:
$ \left[ \tan^{-1}\!\Bigl(\frac{v}{\sqrt{\frac{g}{\gamma}}}\Bigr) \right]_{v_0}^{0}
= \tan^{-1}\!\Bigl(0\Bigr)
- \tan^{-1}\!\Bigl(\frac{v_0}{\sqrt{\frac{g}{\gamma}}}\Bigr). $
Note that $ \tan^{-1}(0) = 0 $. Therefore, the expression becomes:
$ -\, \tan^{-1}\!\Bigl(\sqrt{\frac{\gamma}{g}} \, v_0\Bigr). $
Including the outside factor $ \frac{1}{\gamma} \cdot \frac{1}{\sqrt{\frac{g}{\gamma}}} = \frac{1}{\sqrt{\gamma g}} $ and remembering there is a negative sign on the right side integral, we get:
$ \int_{v_0}^{0} \frac{dv}{g + \gamma v^2} = -\,\frac{1}{\sqrt{\gamma g}} \,\tan^{-1}\!\Bigl(\sqrt{\tfrac{\gamma}{g}} \, v_0\Bigr). $
But this must equal $ - \int_{0}^{t} dt = -\,t. $
7. Solve for the Time $t$
Hence,
$ -\,t = -\,\frac{1}{\sqrt{\gamma g}} \,\tan^{-1}\!\Bigl(\sqrt{\tfrac{\gamma}{g}} \, v_0\Bigr). $
Cancelling the negative signs on both sides gives:
$ t = \frac{1}{\sqrt{\gamma g}} \,\tan^{-1}\!\Bigl(\sqrt{\tfrac{\gamma}{g}} \, V_0\Bigr). $
8. Final Answer
The time taken by the ball to rise to its highest point (zenith) is
$ \displaystyle t = \frac{1}{\sqrt{\gamma g}} \,\tan^{-1}\biggl(\sqrt{\frac{\gamma}{g}} \; V_0\biggr). $
