Question #103772
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
1
Expert's answer
2020-02-28T10:34:02-0500

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 =3×0.23=0.69sec3\times 0.23=0.69sec

So, the ball will reach to the highest point in the time (t1)=0.692=0.345sec(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=ugt1v=u-gt_1

0=ugt10=u-gt_1

u=gt1u=gt_1

So, minimum height thrown by the juggler (h)=ut1gt122(h)=ut_1-\dfrac{gt_1^2}{2}

h=gt12gt122=gt122h=gt_1^2-\dfrac{gt_1^2}{2}=\dfrac{gt_1^2}{2}

h=9.8×0.345×0.3452=0.583mh=\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 ball5×0.23=1.15sec5\times0.23=1.15sec

So, the ball will reach to the highest point in the time (t2)=1.152=0.575(t_2)=\dfrac{1.15}{2}=0.575 sec

at the highest point, velocity (v1)=0(v_1)=0

let the initial velocity (u1)(u_1) and time g is the gravitational acceleration.

v2=u2gt2\Rightarrow v_2=u_2-gt_2

0=u2gt2\Rightarrow 0=u_2-gt_2

u2=gt2u_2=gt_2

Let the minimum height reached by the ball h2=u2t2gt222h_2=u_2t_2-\dfrac{gt_2^2}{2}

h2=gt22gt222=gt222\Rightarrow h_2=gt_2^2-\dfrac{gt_2^2}{2}=\dfrac{gt_2^2}{2}

now substituting the value of t2t_2

h2=9.8×0.575×0.5752=1.62mh_2=\dfrac{9.8\times 0.575\times 0.575}{2}=1.62m


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