Question

Q1)A juggler performs in a room whose ceiling is a height 4.0 mabove 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?

b)What is the time required for the ball to reach the ceiling?

c)At the instant when the first ball is at the ceiling, he throws a second ball upward with two-thirds the initial velocity of the first. How long after the second ball is thrown did the two balls pass each other?

d)At what distance above the juggler's hand do they pass each other?

Answer 1

1) position: y = Ho + Vo*t + ½gt²
0 = 14.6 + v*3.22 - 4.9*(3.22)² → v = 11.2 m/s
is the speed of the scaffolding. Then at impact,
V = v + at = 11.2m/s - 9.8m/s² * 3.22s = -20.9 m/s (that is, down)

When it falls off, it takes t = v / a = 11.2m/s / 9.8m/s² = 1.14 s to reach 0 velocity.
y = v*t + ½at² = 11.2m/s * 1.14s - 4.9m/s² * (1.14s)² = 6.4 m
above the "launch" point. Yes, I'd say he gets the chance.

2) Presumably the ball does not bounce off the ceiling, it reaches zero vertical velocity as it touches.

v = √(2gs)
v = √(2(9.81)4)
v = 8.86 m/s <===ANSWER

t = v/g
t = 8.86 / 9.81
t = 0.903 s <===ANSWER

v₂ = 8.86(2/3)
v₂ = 5.91 m/s

assume hands are origin, up is positive, time from toss of second ball
position equation for first ball

s = s₀ + v₀t + ½gt²

s = 4 + 0t + ½(-9.81)t²

position equation for second ball

s = 0 + 5.91t + ½(-9.81)t²

when they meet, their heights are identical

5.91t + ½(-9.81)t² = 4 + ½(-9.81)t²
5.91t = 4
t = 0.68s <===ANSWER

plug the just found time into either position equation

s = 4 + ½(-9.81)0.68²

s = 1.76 m