A clown in a circus juggles with n balls. He throws each ball vertically upwards with the same speed

Q: Why is the total time of flight $n\tau$ and not $(n-1)\tau$?

If the time of flight was $(n-1)\tau$, the first ball would land exactly when the $n^{\text{th}}$ ball is thrown. But the juggler needs to throw the $n^{\text{th}}$ ball *before* catching the first one to maintain the pattern. The first ball lands at $n\tau$, ready to be thrown again as the $(n+1)^{\text{th}}$ throw.

Q: How did we get the time elapsed for the $i^{\text{th}}$ ball as $(n-i)\tau$?

Think of the throws on a timeline. The $1^{\text{st}}$ ball is thrown at $t=0$, the $2^{\text{nd}}$ at $t=\tau$, and the $i^{\text{th}}$ at $t=(i-1)\tau$. The $n^{\text{th}}$ ball is thrown at $t=(n-1)\tau$. The difference between these two times is $(n-1)\tau - (i-1)\tau = (n-i)\tau$.

Q: Why are the $1^{\text{st}}$ and $3^{\text{rd}}$ balls at the exact same height when $n=4$?

This is the beautiful symmetry of projectile motion! The $1^{\text{st}}$ ball has been in the air for $3\tau$ and is on its way down. The $3^{\text{rd}}$ ball has been in the air for $1\tau$ and is on its way up. Since the total time of flight is $4\tau$, the parabola is symmetric around $t=2\tau$.

Q: Does the mass of the balls affect the juggling pattern?

No! In the absence of air resistance, all objects fall with the same acceleration $g$ regardless of their mass. The kinematics depend entirely on the initial speed and gravity.

Q: Can we solve this using the third equation of motion ($v^2 = u^2 - 2gh$)?

While true, it's not the most efficient path here because we are given information about *time* intervals ($\tau$), not velocities at specific heights. The second equation ($h = ut - \frac{1}{2}gt^2$) directly connects height with time.

Question

A clown in a circus juggles with $n$ balls. He throws each ball vertically upwards with the same speed at equal time intervals $	au$. Denote acceleration of free fall by $g$.
(a) Find expressions for the speed of projection and height of the $i^{	ext{th}}$ ball above his hand when he throws the $n^{	ext{th}}$ ball.
If he uses $n = 4$ balls, distance between the second and third ball is $d = 50	ext{ cm}$ at the instant the fourth ball is projected.
(b) Where is the first ball, when the juggler throws the fourth ball?
(c) What is maximum height attained by each ball above the hands of the juggler?

Sigma Percentile · Accepted Answer

### The Art and Physics of Juggling

Imagine a clown in a circus, effortlessly keeping multiple balls in the air. It looks like magic, but it is pure, unadulterated physics. To maintain a continuous, rhythmic loop, the juggler must rely on precise timing and the unforgiving laws of gravity.

Let's break down the mechanics. If the clown is juggling $n$ balls and throws them at equal time intervals $	au$, the first ball must return to his hand exactly when it is time to make the $(n+1)^{	ext{th}}$ throw. This means the total time of flight $T$ for any single ball is simply the number of balls multiplied by the time interval:

$$T = n	au$$

### Finding the Speed of Projection

Using basic kinematics, we know that the time of flight for an object thrown vertically upwards with initial speed $u$ is given by $T = \frac{2u}{g}$. By equating our two expressions for $T$, we can easily find the required speed of projection:

$$\frac{2u}{g} = n	au \implies u = \frac{1}{2}gn	au$$

This elegant equation tells us that the speed required is directly proportional to the number of balls and the time interval between throws.

### The Master Equation for Height

Now, let's freeze time at the exact instant the $n^{	ext{th}}$ ball is thrown. We want to find the height $h_i$ of the $i^{	ext{th}}$ ball.

First, we need to know how long the $i^{	ext{th}}$ ball has been in the air. If the $1^{	ext{st}}$ ball is thrown at $t=0$, the $i^{	ext{th}}$ ball is thrown at $t=(i-1)	au$, and the $n^{	ext{th}}$ ball is thrown at $t=(n-1)	au$. The time elapsed for the $i^{	ext{th}}$ ball is the difference between these two times:

$$t_i = (n-1)	au - (i-1)	au = (n-i)	au$$

We can now substitute this elapsed time and our initial speed $u$ into the second equation of motion, $h = ut - \frac{1}{2}gt^2$:

$$h_i = \left(\frac{1}{2}gn	auight) (n-i)	au - \frac{1}{2} g ((n-i)	au)^2$$

Factoring out the common terms, we arrive at a beautifully symmetric expression for the height:

$$h_i = \frac{1}{2}g	au^2 [n(n-i) - (n-i)^2]$$

$$h_i = \frac{1}{2}g	au^2 (n-i)[n - (n-i)]$$

**$$h_i = \frac{1}{2}g	au^2 i(n-i)$$**

### Solving the Specific Case ($n=4$)

The problem gives us a specific scenario: the clown is juggling $4$ balls, and the distance between the $2^{	ext{nd}}$ and $3^{	ext{rd}}$ balls is $50	ext{ cm}$ ($0.5	ext{ m}$) at the instant the $4^{	ext{th}}$ ball is thrown.

Let's plug $i=2$ and $i=3$ into our master equation:

$$h_2 = \frac{1}{2}g	au^2 (2)(4-2) = 2g	au^2$$

$$h_3 = \frac{1}{2}g	au^2 (3)(4-3) = 1.5g	au^2$$

The difference in their heights is given as $0.5	ext{ m}$:

$$h_2 - h_3 = 2g	au^2 - 1.5g	au^2 = 0.5g	au^2$$

$$0.5g	au^2 = 0.5 \implies \mathbf{g	au^2 = 1}$$

This simple numerical value is the key to unlocking the rest of the problem.

### Final Calculations

**Where is the first ball?**
We substitute $i=1$ into our height formula:

$$h_1 = \frac{1}{2}g	au^2 (1)(4-1) = 1.5g	au^2$$

Since $g	au^2 = 1$, the first ball is at a height of **$1.5	ext{ m}$**.

**What is the maximum height?**
The maximum height $H_{	ext{max}}$ is given by $\frac{u^2}{2g}$. Substituting our expression for $u$ (with $n=4$):

$$H_{	ext{max}} = \frac{(2g	au)^2}{2g} = 2g	au^2$$

Again, since $g	au^2 = 1$, the maximum height is **$2.0	ext{ m}$**.

Notice the incredible symmetry here: the $1^{	ext{st}}$ ball and the $3^{	ext{rd}}$ ball are at the exact same height ($1.5	ext{ m}$). The $3^{	ext{rd}}$ ball is on its way up, while the $1^{	ext{st}}$ ball is on its way down. This perfect parabolic symmetry is what makes juggling visually mesmerizing!

Visualized Solution

$T = n τ$

$u = \frac{1}{2} g n τ$

$t_{i} = (n - i) τ$

$h_{i} = u t_{i} - \frac{1}{2} g t_{i}^{2}$

$h_{i} = \frac{1}{2} g τ^{2} i (n - i)$

$n = 4, h_{2} - h_{3} = 0.5 m$

$g τ^{2} = 1$

$h_{1} = 1.5 m$

$H_{max} = 2.0 m$

Symmetry in Juggling

The Sigma Insight: Motion in a Straight Line

The Art and Physics of Juggling

Finding the Speed of Projection

The Master Equation for Height

Solving the Specific Case ( $n = 4$ )

Final Calculations

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u=21​gnτ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ti​=(n−i)τ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } hi​=uti​−21​gti2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } hi​=21​gτ2i(n−i)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=4,h2​−h3​=0.5 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } gτ2=1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h1​=1.5 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hmax​=2.0 m