Answer to Question #103772 in Classical Mechanics for Ayatollah

Question

1. (Part a) A juggler throws balls almost vertically upwards, with time delta t = 0.23 s between each throw. (Yes, pretty amazing.) The balls are all thrown to the same height, and in repeated succession (e.g. 1, 2, 3, 1, 2, 3 etc.). Let's simplify: neglect air resistance, neglect the time between catching and throwing a ball and assume that balls are thrown and caught at the same height. Calculate the minimum height he must throw the balls if he is to juggle

3 balls? _____ m

(Part b) Using the information from question (part a) above: Calculate the minimum height he must throw the balls if he is to juggle

5 balls? _____ m

Answer 1

As per the question,

The time between each throw=0.23 sec

a)

In case of three ball, the first ball is about to reach his hand, then the time of journey of each ball =$3\times 0.23=0.69sec$

So, the ball will reach to the highest point in the time $(t_1)=\dfrac{0.69}{2}=0.345sec$

at the top point the final velocity(v) =0

let the initial velocity = u

let g is the gravitational acceleration

$v=u-gt_1$

$0=u-gt_1$

$u=gt_1$

So, minimum height thrown by the juggler $(h)=ut_1-\dfrac{gt_1^2}{2}$

$h=gt_1^2-\dfrac{gt_1^2}{2}=\dfrac{gt_1^2}{2}$

$h=\dfrac{9.8\times0.345\times 0.345}{2}=0.583m$

b)

In case of five ball, the first ball is about to reach his hand, then the time of journey of each ball$5\times0.23=1.15sec$

So, the ball will reach to the highest point in the time $(t_2)=\dfrac{1.15}{2}=0.575$ sec

at the highest point, velocity $(v_1)=0$

let the initial velocity $(u_1)$ and time g is the gravitational acceleration.

$\Rightarrow v_2=u_2-gt_2$

$\Rightarrow 0=u_2-gt_2$

$u_2=gt_2$

Let the minimum height reached by the ball $h_2=u_2t_2-\dfrac{gt_2^2}{2}$

$\Rightarrow h_2=gt_2^2-\dfrac{gt_2^2}{2}=\dfrac{gt_2^2}{2}$

now substituting the value of $t_2$

$h_2=\dfrac{9.8\times 0.575\times 0.575}{2}=1.62m$