A stone projected from edge A of a high cliff strikes the ground at point C moving almost vertically

Q: Why does the stone fall almost vertically at the end?

Air resistance continuously removes kinetic energy. The horizontal velocity decays exponentially as $v_x = u_x e^{-\alpha t}$. If the cliff is high enough, $v_x$ becomes negligible before it hits the ground, making the final trajectory purely vertical.

Q: Why is the time of ascent different from the time of descent?

During ascent, gravity and air resistance act in the same downward direction, causing a large retardation. During descent, air resistance acts upwards against gravity, resulting in a smaller net acceleration. Thus, it takes longer to come down.

Q: How did we know to integrate the acceleration equation directly?

Solving the full differential equation for $y(t)$ is tedious. However, recognizing that the integral of velocity over time is displacement ($\int v_y dt = \Delta y$) allows us to bypass finding the explicit function $y(t)$ and directly relate velocities, times, and height.

Q: Why is the displacement $-h$ for the downward journey?

We defined the upward direction as positive. The stone moves from the maximum height (coordinate $h$) back to the level AB (coordinate $0$). The change in position is final minus initial, which is $0 - h = -h$.

Q: Is the horizontal range $R$ exactly equal to $u_x / \alpha$?

Not exactly. The true range is $x(t) = \frac{u_x}{\alpha}(1 - e^{-\alpha t})$. However, because the stone hits the ground 'almost vertically', $e^{-\alpha t}$ is extremely small, making $R \approx u_x / \alpha$ a highly accurate approximation.

Question

A stone projected from edge A of a high cliff strikes the ground at point C moving almost vertically. Reason for this strange behavior is air resistance that is proportional to the speed of the stone. The points A and B on the trajectory are in the same horizontal level. Time taken by the stone in its upward and downward motions above the level AB differ by $\Delta t$ and moduli of vertical component of velocities at points A and B differ by $\Delta v_y$. Horizontal component of velocity at point A is $u_x$ and horizontal displacement of the stone from A to C is $R$. Denoting acceleration due to gravity by $g$, find suitable expression for the maximum height of the stone above the horizontal level AB.

Sigma Percentile · Accepted Answer

### The Asymmetry of Real-World Projectiles

When you throw a stone in a vacuum, it traces a perfect, symmetric parabola. But introduce air resistance, and the physics becomes beautifully complex. The drag force, proportional to velocity ($\vec{F}_{drag} = -k \vec{v}$), continuously saps kinetic energy from the stone. This means the acceleration is no longer just gravity; it has a horizontal component $a_x = -\alpha v_x$ and a vertical component $a_y = -g - \alpha v_y$, where $\alpha = k/m$.

### Decoupling the Dimensions

The brilliance of this problem lies in how the horizontal and vertical motions, while independent in their differential equations, are intimately linked by time. Let's look at the horizontal motion first. The equation $\frac{dv_x}{dt} = -\alpha v_x$ tells us that the horizontal velocity decays exponentially: $v_x(t) = u_x e^{-\alpha t}$.

Integrating this gives the horizontal position $x(t) = \frac{u_x}{\alpha} (1 - e^{-\alpha t})$.

### The Horizontal Limit

The question provides a massive clue: the stone strikes the ground "almost vertically." This implies that by the time it reaches point C, its horizontal velocity has decayed to nearly zero ($v_x \approx 0$). Mathematically, this happens as $t 	o \infty$.

In this limit, the horizontal displacement reaches its maximum possible value, which is exactly the range $R$. Therefore, $R \approx \frac{u_x}{\alpha}$. This elegant limit allows us to express the unknown drag constant $\alpha$ entirely in terms of known initial conditions: $\alpha \approx \frac{u_x}{R}$.

### The Vertical Integration Trick

Now, we face the vertical motion: $\frac{dv_y}{dt} = -g - \alpha v_y$. Solving this for $y(t)$ explicitly is a nightmare of exponential functions. But we don't need to! We can use a powerful integration trick. We know that the integral of velocity over time is simply displacement: $\int v_y dt = \Delta y$.

Let's integrate the acceleration equation over the upward journey (from $t=0$ to $t=t_1$). The velocity goes from $u_y$ to $0$, and the displacement is the maximum height $h$:

$\int_{u_y}^{0} dv_y = \int_{0}^{t_1} (-g - \alpha v_y) dt$

$-u_y = -g t_1 - \alpha h \implies u_y = g t_1 + \alpha h$

We do the exact same thing for the downward journey (from $t=t_1$ to $t=t_1+t_2$). The velocity goes from $0$ to $-v_B$, and the displacement is $-h$ (since it's falling back to the same level):

$\int_{0}^{-v_B} dv_y = \int_{t_1}^{t_1+t_2} (-g - \alpha v_y) dt$

$-v_B = -g t_2 - \alpha (-h) \implies v_B = g t_2 - \alpha h$

### Synthesizing the Final Result

We now have two clean algebraic equations. By subtracting the downward equation from the upward one, we can eliminate the individual times and velocities, replacing them with the differences given in the problem:

$u_y - v_B = g(t_1 - t_2) + 2\alpha h$

We are given that the difference in velocity moduli is $\Delta v_y = u_y - v_B$, and the difference in time is $\Delta t = t_2 - t_1$. Substituting these in:

$\Delta v_y = -g \Delta t + 2\alpha h$

Rearranging for the maximum height $h$, we get $h = \frac{\Delta v_y + g \Delta t}{2\alpha}$. Finally, substituting our previously found $\alpha \approx \frac{u_x}{R}$, we arrive at the magnificent final expression:

**$h \approx \frac{R}{u_x} \left( \frac{\Delta v_y + g \Delta t}{2} ight)$**

Visualized Solution

$Understanding the Drag Force$

$Horizontal Motion and Terminal State$

$Finding the Drag Constant α$

$Vertical Motion: Upward Journey$

$Vertical Motion: Downward Journey$

$Relating Velocities and Times$

$Final Expression for h$

The Sigma Insight: Projectile Motion

The Asymmetry of Real-World Projectiles

Decoupling the Dimensions

The Horizontal Limit

The Vertical Integration Trick

Synthesizing the Final Result

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Horizontal Motion and Terminal State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Drag Constant α

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

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

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

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