Question 12: Juggler with n Balls
Solution
(a) Expressions for Speed and Height
Let $\tau$ be the time interval between throws. Since there are $n$ balls, and the $n$-th ball is just being thrown while the previous $(n-1)$ balls are in the air, the first ball thrown must be landing or completing its cycle. For continuous juggling, the total time of flight $T$ for one ball must accommodate the $n$ intervals required to cycle through all balls.
Condition: $T = n\tau$.
Using $T = \frac{2u}{g}$:
$$ \frac{2u}{g} = n\tau \implies u = \frac{gn\tau}{2} $$
For the $i$-th ball:
The $n$-th ball is thrown at $t=0$. The $i$-th ball was thrown $(n-i)$ intervals ago.
Time elapsed for $i$-th ball: $t_i = (n-i)\tau$.
Height of $i$-th ball:
$$ h_i = u t_i – \frac{1}{2}g t_i^2 $$
$$ h_i = \left( \frac{gn\tau}{2} \right)(n-i)\tau – \frac{1}{2}g \left( (n-i)\tau \right)^2 $$
$$ h_i = \frac{1}{2}g\tau^2 (n-i) [ n – (n-i) ] $$
$$ h_i = \frac{1}{2}g\tau^2 i(n-i) $$
(b) Case $n=4$: Position of First Ball
Given distance between 2nd and 3rd ball is $d = 50 \text{ cm} = 0.5 \text{ m}$.
Time elapsed for 2nd ball ($i=2$): $t_2 = (4-2)\tau = 2\tau$.
Time elapsed for 3rd ball ($i=3$): $t_3 = (4-3)\tau = \tau$.
Heights:
$$ h_2 = \frac{1}{2}g\tau^2 (2)(4-2) = \frac{1}{2}g\tau^2 (4) = 2g\tau^2 $$
$$ h_3 = \frac{1}{2}g\tau^2 (3)(4-3) = \frac{1}{2}g\tau^2 (3) = 1.5g\tau^2 $$
Difference:
$$ |h_2 – h_3| = 0.5g\tau^2 $$
Given $|h_2 – h_3| = 0.5$ m:
$$ 0.5g\tau^2 = 0.5 \implies g\tau^2 = 1 $$
Position of 1st ball ($i=1$) when 4th is thrown:
$$ h_1 = \frac{1}{2}g\tau^2 (1)(4-1) = \frac{1}{2}(1)(3) = 1.5 \text{ m} $$
Answer: 1.5 m
(c) Maximum Height
Max height $H = \frac{u^2}{2g}$.
Substitute $u = \frac{gn\tau}{2}$:
$$ H = \frac{1}{2g} \left( \frac{gn\tau}{2} \right)^2 = \frac{g^2 n^2 \tau^2}{8g} = \frac{g n^2 \tau^2}{8} $$
Using $n=4$ and $g\tau^2 = 1$:
$$ H = \frac{4^2 (1)}{8} = \frac{16}{8} = 2.0 \text{ m} $$
Answer: 2.0 m