Problem 4: Juggling Act. A juggler performs in a room whose ceiling is 3.0m abov

Question

Problem 4: Juggling Act. A juggler performs in a room whose ceiling is 3.0m above the level of his hands. He throws a ball upward so that it just reaches the ceiling. a) What is the initial velocity of the ball? What is the time required to for the ball to reach the ceiling? b) is at the ceiling, the juggler throw thirds the initial velocity of the first. c) How long after the second ball is thrown do the two balls pass each other? d) At what distance above the juggler's hand do they pass each other?

Explanation / Answer

Given

Ceilling is at a height is h = 3 m

when the ball was thrown with initial velocity so that it can reach maximum of height 3 m  

a)

so v^2 - u^2 = 2gH

h = u^2/2g at maximum height final velocity is zero

3 = u^2/(2*9.8)

u = 7.668m/s

b)

time required to reach the ceiling is  

v = u-gt

t = u/g

t = 7.668/9.8 s

t = 0.78245 s

c)

the initial velocity of the ball b1 thrown upward is u = (2/3)(7.668) m/s = 5.112 m/s

the second ball b2 dropped from the ceiling of height 3 m

here if b1 covers a distance of x in time t so that b2 can cover a distance of 3-x m in the same time t

writing the equations of motions for b1,b2

for b1

x = ut-0.5*gt^2

for b2

3-x = 0.5*g*t^2

adding the equations we get

x = 5.112*t-0.5*9.8*t^2

3-x = 0.5*9.8*t^2

--------------------------------------------

3 = 5.112*t ==> t = 0.5869 s

--------------------------------------------

so the two balls can pass each other after the second ball was thrown is 0.5869 s

d)

substituting the value of t isn equation x = ut-0.5*gt^2

x = 5.112*t-0.5*9.8*t^2

x = 5.112*0.5869-0.5*9.8*0.5869^2 = 1.31242 m

the balls can pass each other at 1/31242 m

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