A ball is thrown upward with an initial velocity v0 from the surface of the earth. The motion of the

Q: Why are both forces negative in the net force equation?

Because we chose the upward direction as positive. Since gravity and drag both pull the ball downwards, they are acting in the negative direction.

Q: Why did we use $dv/dt$ instead of $v \frac{dv}{dx}$?

The question specifically asks for the *time* taken to reach the zenith. Using $dv/dt$ introduces time into our equation. If we needed to find the maximum height, we would use $v \frac{dv}{dx}$ to introduce position.

Q: What happens to the drag force when the ball falls back down?

When the ball falls, its velocity is directed downwards. The drag force always opposes motion, so it will act *upwards*, while gravity still acts downwards. The net force equation would change to $F_{\text{net}} = mg - m\gamma v^2$ (taking downward as positive).

Q: How do we know the velocity is zero at the zenith?

The zenith is the highest point of the trajectory. For the ball to stop going up and start coming down, its velocity must momentarily become zero at that exact peak.

Q: What if there was no drag force ($\gamma = 0$)?

If $\gamma \to 0$, we can use the small angle approximation $\tan^{-1}(x) \approx x$. The expression simplifies to $t = \frac{1}{\sqrt{\gamma g}} \left(\sqrt{\frac{\gamma}{g}} v_0\right) = \frac{v_0}{g}$, which is the classic kinematic result you learn in basic projectile 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 $m\gamma v^2$ (where, $m$ is 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

Options:

(A) $\frac{1}{\sqrt{2\gamma g}} 	an^{-1}\left(\sqrt{\frac{2\gamma}{g}} v_0ight)$

(B) $\frac{1}{\sqrt{\gamma g}} 	an^{-1}\left(\sqrt{\frac{\gamma}{g}} v_0ight)$

(C) $\frac{1}{\sqrt{\gamma g}} \sin^{-1}\left(\sqrt{\frac{\gamma}{g}} v_0ight)$

(D) $\frac{1}{\sqrt{\gamma g}} \ln\left(1 + \sqrt{\frac{\gamma}{g}} v_0ight)$

Sigma Percentile · Accepted Answer

### The Setup: Gravity and Drag

Imagine you are standing on the surface of the Earth, and you throw a ball straight up into the sky with an initial velocity $v_0$. If we lived in a perfect vacuum, gravity would be the only force acting on the ball. But in the real world, the air pushes back.

As the ball moves upward with velocity $v$, it experiences two distinct forces trying to pull it back down. First, there is the relentless pull of gravity, $mg$. Second, there is the aerodynamic drag force, which the problem defines as $m\gamma v^2$. Because both of these forces oppose the upward motion of the ball, they both act downwards.

### The Mathematical Translation

To find out how long it takes for the ball to reach its highest point (the zenith), we need to translate this physical reality into mathematics using Newton's Second Law.

Taking the upward direction as positive, the net force on the ball is:
$$F_{	ext{net}} = -mg - m\gamma v^2$$

Since $F_{	ext{net}} = ma$, we can write:
$$ma = -m(g + \gamma v^2)$$

The mass $m$ beautifully cancels out from both sides, leaving us with the net acceleration of the ball:
$$a = -(g + \gamma v^2)$$

Now, we need to connect this acceleration to time. We know from kinematics that acceleration is the rate of change of velocity, so we can substitute $a = \frac{dv}{dt}$:
$$\frac{dv}{dt} = -(g + \gamma v^2)$$

### The Art of Integration

We now have a first-order differential equation. To solve it, we must separate the variables, bringing all the $v$ terms to one side and the $dt$ to the other:
$$\frac{dv}{g + \gamma v^2} = -dt$$

It's time to integrate. We must carefully define our limits based on the physical journey of the ball. At the moment of launch ($t = 0$), the velocity is $v_0$. At the zenith, the ball momentarily stops before falling back down, so at time $t$, the velocity is $0$.

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

To evaluate the left integral, let's factor out $\gamma$ from the denominator to match a standard integration form:
$$\frac{1}{\gamma} \int_{v_0}^{0} \frac{dv}{\frac{g}{\gamma} + v^2} = -t$$

This perfectly matches the standard integral $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}\left(\frac{x}{a}ight)$, where our constant $a$ is $\sqrt{\frac{g}{\gamma}}$.

### The Final Destination

Applying the integration formula, we get:
$$\frac{1}{\gamma} \left[ \frac{1}{\sqrt{\frac{g}{\gamma}}} 	an^{-1}\left(\frac{v}{\sqrt{\frac{g}{\gamma}}}ight) ight]_{v_0}^{0} = -t$$

Simplifying the constants outside the bracket gives $\frac{1}{\sqrt{\gamma g}}$. Now, we substitute our upper and lower limits:
$$\frac{1}{\sqrt{\gamma g}} \left[ 	an^{-1}(0) - 	an^{-1}\left(\sqrt{\frac{\gamma}{g}} v_0ight) ight] = -t$$

Since $	an^{-1}(0) = 0$, the equation simplifies to:
$$-\frac{1}{\sqrt{\gamma g}} 	an^{-1}\left(\sqrt{\frac{\gamma}{g}} v_0ight) = -t$$

The negative signs on both sides cancel out, revealing the final, elegant expression for the time taken to reach the zenith:
$$t = \frac{1}{\sqrt{\gamma g}} 	an^{-1}\left(\sqrt{\frac{\gamma}{g}} v_0ight)$$

Notice how the presence of the velocity-squared drag force transforms a simple linear time equation into one governed by the inverse tangent function. Physics is beautiful!

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Visualized Solution

$Free Body Diagram$

$Newton’s Second Law$

$Kinematic Equation$

$Separation of Variables$

$Setting up the Integral$

$Integration$

$Evaluating Limits$

$Final Answer$

$The Way Forward$

The Sigma Insight: Newton's Laws of Motion

The Setup

The Mathematical Translation

The Art of Integration

The Final Destination

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Body Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Newton’s Second Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinematic Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Separation of Variables

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Integral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Limits

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward