Two balls are dropped from the top of a cliff at a time interval delta t = 2 s. The first ball hits

Q: Why do both balls take the same time to fall to height h? Ball 1 was dropped earlier!

The time it takes to fall a certain distance from rest depends only on gravity and that distance ($t = \sqrt{2d/g}$). Since both balls are dropped from rest and fall the same distance ($H-h$) to reach the collision point, their travel times for that specific segment must be identical, regardless of *when* they were dropped.

Q: How do we know Ball 1 spends exactly $\Delta t$ seconds below height h?

Let's say it takes 5 seconds to fall to height h. Ball 1 reaches height h on its way down at $t=5$s. Ball 2 is dropped at $t=2$s, so it reaches height h at $t=2+5=7$s. For them to collide, Ball 1 must also be at height h at $t=7$s. The difference between $t=7$s and $t=5$s is exactly the 2-second delay ($\Delta t$)!

Q: Why is the time to go down from h equal to the time to go up to h?

The problem states the bounce is perfectly elastic, meaning the ball doesn't lose any kinetic energy. It rebounds with the exact same speed it hit the ground with. In kinematics, motion under constant acceleration is symmetric if initial and final speeds at a given height are the same.

Q: Why did we use $v = \sqrt{2g(H-h)}$ for the initial velocity in the second part?

We split Ball 1's journey into two parts. The first part is falling from the top of the cliff ($H$) to height $h$. It starts from rest, so its velocity after falling that distance ($H-h$) is given by $v^2 = u^2 + 2as$, which becomes $v = \sqrt{2g(H-h)}$.

Q: What if the balls had different masses? Would the answer change?

No! In the absence of air resistance, all objects fall at the same rate regardless of their mass, due to the equivalence of gravitational and inertial mass. The kinematic equations we used do not depend on mass at all.

Question

Two balls are dropped from the top of a cliff at a time interval $\Delta t = 2	ext{ s}$. The first ball hits the ground, rebounds elastically (reversing direction instantly without losing speed), and collides with the second ball at a height $h = 55	ext{ m}$ above the ground. How high is the top of the cliff?

Sigma Percentile · Accepted Answer

### The Symphony of Falling Balls: A Tale of Relative Time

Imagine standing at the edge of a towering cliff. You hold a ball over the edge and let it go. Exactly two seconds later, you drop a second ball. The first ball plummets to the ground, bounces back up with perfect elasticity, and smacks right into the second ball at a height of $55	ext{ m}$.

Our mission is to find the total height of this cliff, $H$. At first glance, this seems like a chaotic system of multiple moving parts. But if we look closely at the timeline, a beautiful symmetry emerges.

### Analyzing the Timeline

Let's break down the journey. Both balls are dropped from rest from the exact same height. Therefore, the time it takes for either ball to fall from the top of the cliff down to the collision height $h$ is identical. Let's call this time $t_0$.

Because the second ball was dropped $\Delta t = 2	ext{ s}$ later, it arrives at the collision point at time $t = t_0 + \Delta t$.

Now, here is the crucial insight: The first ball passed this exact height $h$ on its way down at time $t_0$. Yet, it meets the second ball at time $t_0 + \Delta t$. This means the first ball spent exactly $\Delta t$ seconds traveling from height $h$, down to the ground, and bouncing back up to height $h$!

### The Power of Symmetry

The problem states that the bounce is perfectly elastic. This means there is no loss of kinetic energy, and the motion is perfectly symmetric. The time it takes to fall from height $h$ to the ground is exactly equal to the time it takes to rise back up from the ground to height $h$.

Since the total time for this round trip is $\Delta t$, the downward journey from height $h$ to the ground took exactly half of that time, which is $\frac{\Delta t}{2}$.

### The Master Equation

Now we can use our standard kinematics. Let's focus solely on the downward segment from height $h$ to the ground. Let the velocity of the ball as it passes height $h$ be $v$. Using the second equation of motion, $s = ut + \frac{1}{2}at^2$, we get:

$$h = v \left( \frac{\Delta t}{2} ight) + \frac{1}{2} g \left( \frac{\Delta t}{2} ight)^2$$

But what is this velocity $v$? The ball was dropped from rest from the top of the cliff, falling a distance of $(H - h)$. Using the third equation of motion, $v^2 = u^2 + 2as$, we find that $v = \sqrt{2g(H - h)}$.

Substituting this back into our height equation gives us our master equation:

$$h = \sqrt{2g(H - h)} \left( \frac{\Delta t}{2} ight) + \frac{1}{8} g (\Delta t)^2$$

### Final Calculation

Now, let's plug in the numbers given in the problem: $h = 55	ext{ m}$, $\Delta t = 2	ext{ s}$, and $g = 10	ext{ m/s}^2$.

$$55 = \sqrt{2(10)(H - 55)} \left( \frac{2}{2} ight) + \frac{1}{8} (10) (2)^2$$

Simplifying the terms, the time fraction becomes $1$, and the gravity term $\frac{10 	imes 4}{8}$ simplifies neatly to $5$:

$$55 = \sqrt{20(H - 55)} + 5$$

Subtracting $5$ from both sides isolates the radical:

$$50 = \sqrt{20(H - 55)}$$

To get rid of the square root, we square both sides. $50^2$ is $2500$:

$$2500 = 20(H - 55)$$

Dividing by $20$ gives:

$$125 = H - 55$$

Finally, adding $55$ to $125$, we find the total height of the cliff:

**$$H = 180	ext{ m}$$**

This is a magnificent problem that elegantly tests our understanding of relative time, symmetric motion, and the independence of falling bodies.

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Setup}

\text{Analyzing the Time of Flight}

\text{The Journey of Ball 1}

\text{Symmetry of Elastic Bounce}

\text{Kinematics of the Downward Journey}

\text{Velocity at Height } h

\text{Substituting the Values}

\text{Simplifying the Equation}

\text{Isolating the Radical}

\text{Squaring Both Sides}

\text{Final Height of the Cliff}

\text{The Way Forward}

The Sigma Insight: Motion in a Straight Line

The Symphony of Falling Balls

Analyzing the Timeline

The Power of Symmetry

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Analyzing the Time of Flight} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{The Journey of Ball 1} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Symmetry of Elastic Bounce} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Kinematics of the Downward Journey} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Velocity at Height } h .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Final Height of the Cliff} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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