A body falling freely from a given height H hits an inclined plane in its path at a height h. As a r

Q: Why does the horizontal velocity after impact not affect the time to reach the ground?

In projectile motion, the horizontal and vertical motions are completely independent. The time of flight is determined solely by the initial vertical velocity and the vertical distance to fall. Since the vertical velocity is reset to zero after the impact, the horizontal speed has no effect on the time $t_2$.

Q: What happens to the kinetic energy during the impact?

The problem implies that the direction of velocity changes to horizontal, but it doesn't explicitly state if the collision is elastic. However, for calculating the time of flight, the magnitude of the horizontal velocity doesn't matter. We only care that the vertical component becomes zero.

Q: Why do we differentiate with respect to $h$ and not $H$?

$H$ is the total given height, which is a constant. The variable we can control or choose is $h$, the height at which the inclined plane is placed. Therefore, we optimize the function with respect to the variable $h$.

Q: Could we have solved this without calculus?

Yes! You can use the algebraic inequality AM-GM or Cauchy-Schwarz. For instance, by Cauchy-Schwarz, $(1 \cdot \sqrt{H-h} + 1 \cdot \sqrt{h})^2 \le (1^2 + 1^2)(H-h + h) = 2H$. The maximum value is $\sqrt{2H}$, which occurs when $\sqrt{H-h} / 1 = \sqrt{h} / 1$, meaning $H-h = h$, or $h = H/2$.

Q: What if the velocity didn't become completely horizontal?

If the velocity had a downward vertical component after impact, the time $t_2$ would be shorter. If it had an upward component, it would be longer. The problem specifically simplifies this by resetting the vertical velocity to zero.

Question

A body falling freely from a given height $H$ hits an inclined plane in its path at a height $h$. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of $(h / H)$, the body will take maximum time to reach the ground?

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Falls

Imagine dropping a ball from a tall building of height $H$. On its way down, it strikes a slanted roof at a height $h$ above the ground. This collision is special: it perfectly redirects the ball's velocity to be purely horizontal.

This single event splits the ball's journey into two completely independent phases. The first phase is a pure vertical free fall from the top down to the roof. The second phase is a horizontal projectile motion from the roof down to the ground.

Our mission? To find the exact height $h$ where we should place this roof to keep the ball in the air for the absolute maximum amount of time.

### Phase 1: The Free Fall

Let's analyze the first part of the journey. The ball starts from rest, so its initial velocity is zero. It falls from the initial height $H$ down to the roof at height $h$. Therefore, the vertical distance covered is $H - h$.

Using the second equation of motion, $s = ut + \frac{1}{2}at^2$, we can find the time taken for this first leg, which we will call $t_1$.

$$H - h = 0 + \frac{1}{2}gt_1^2$$

Rearranging this to solve for $t_1$, we get:

$$t_1 = \sqrt{\frac{2(H - h)}{g}}$$

### Phase 2: The Horizontal Projectile

Now, the ball hits the roof. The problem states that its velocity becomes entirely horizontal. **This is the crucial physical insight.** Because the velocity is purely horizontal, the vertical component of the velocity is instantaneously reset to zero!

For the second part of the journey, the ball must fall the remaining vertical distance $h$. Since its initial vertical velocity for this phase is zero, the horizontal motion has absolutely no effect on how long it takes to hit the ground.

We apply the same kinematic equation for the vertical direction to find the time $t_2$:

$$h = 0 + \frac{1}{2}gt_2^2$$

Solving for $t_2$, we find:

$$t_2 = \sqrt{\frac{2h}{g}}$$

### The Master Equation: Total Time

The total time the ball spends in the air, $t$, is simply the sum of the time spent in Phase 1 and Phase 2.

$$t = t_1 + t_2$$

Substituting our expressions for $t_1$ and $t_2$, we get the master equation for the total time:

$$t = \sqrt{\frac{2(H - h)}{g}} + \sqrt{\frac{2h}{g}}$$

We can factor out the common constant $\sqrt{\frac{2}{g}}$ to make it cleaner:

$$t = \sqrt{\frac{2}{g}} \left( \sqrt{H - h} + \sqrt{h} \right)$$

### The Calculus of Optimization

We want to maximize this total time $t$. In calculus, to find the maximum value of a function, we take its derivative with respect to the variable we can change—in this case, $h$—and set it equal to zero.

Let's differentiate $t$ with respect to $h$:

$$\frac{dt}{dh} = \sqrt{\frac{2}{g}} \left[ \frac{d}{dh}(\sqrt{H - h}) + \frac{d}{dh}(\sqrt{h}) \right] = 0$$

Applying the chain rule to the first term and the power rule to the second, we get:

$$\frac{1}{2\sqrt{H - h}} \cdot (-1) + \frac{1}{2\sqrt{h}} = 0$$

**Watch out for that minus sign!** It comes from the derivative of the inner function $(H - h)$.

### The Final Calculation

Now, we just need to solve this algebraic equation for $h$. Let's move the negative term to the right side:

$$\frac{1}{2\sqrt{h}} = \frac{1}{2\sqrt{H - h}}$$

We can cancel the $2$ in the denominators. Then, squaring both sides to eliminate the square roots, we arrive at a beautifully simple relationship:

$$h = H - h$$

Adding $h$ to both sides gives:

$$2h = H$$

Finally, dividing by $H$ and $2$, we find the golden ratio:

$$\frac{h}{H} = \frac{1}{2}$$

**The physical meaning is profound:** To maximize the total time of flight, the inclined plane must be placed exactly halfway between the release point and the ground. This perfectly balances the time spent accelerating in the first half with the time spent falling from rest in the second half!

Enter Numerical Value:

Visualized Solution

Visualizing the Two-Part Journey

Time of Free Fall ( $t_{1}$ )

Calculating $t_{1}$

The Impact and Horizontal Velocity

Time of Projectile Motion ( $t_{2}$ )

Total Time of Flight ( $t$ )

Maximizing the Total Time

Differentiating the Time Function

Solving for $h$

The Final Ratio

The Sigma Insight: Projectile Motion

The Setup

Phase 1

Phase 2

The Master Equation

The Calculus of Optimization

The Final Calculation

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