Question

A juggler performs in a room whose ceiling is 3.0 m above the level of his hands. He throws a ball upward so that it reaches the ceiling. (a). What is the initial velocity of the ball? (b). What is the time required for the ball to reach the ceiling? At the instant when the first ball is at the ceiling, the juggler throws a second ball upward with 2/3 the initial velocity of the first. (c). How long after the second ball is thrown did the two balls pass each other? (d). At what distance above the juggle's hand do they pass each other?

Answer 1

Answer:

ans. below

Explanation:

let g = 10m/s²

o = u - 10t , t = u/10

ut - 5t² = 15

u²/10 - u²/20 = u²/20 = 15 , u² = 300 , u = 10√3 m/s

a) u = 10√3 m/s

b) time t = u/10 = √3 sec

c) for the second ball u = 20/√3m/s

let the two balls meet after t sec. at height H above the  the juggle's hand .

3 - H = 5t²

H = 20t/√3 - 5t²

3 = 20t/√3 - 5t²

5t² - 20t/√3 + 3 = 0

t² -  4t/√3 + 0.6 = 0

t = { 4/√3 ± √ 16/3 - 2.4}/2 = { 4/√3 ± √ 8.8/3}/2 = { 4 ± √ 8.8}/2√3

t = { 2 ± √2.2}/√3 = { 2 - √2.2}/√3 sec.

d) H = 3 - 5t² = 3 - 5  { 2 - √2.2}²/3 = 2.555m