1. May 12, 2007

### budynavan

Hi everyone, I have a difficulty of the following Dynamics (kinetics) problems (from Hibbeler’s Eng Mech Dynamics, 11E book)

PROBLEMS 12-26

Ball A is released from rest at a height of 40 ft at the same time that a second ball B is thrown upward 5 ft from the ground. If the balls pass one another at a height of 20 ft, determine the speed at which ball B was thrown upward.

Given:

h1 = 40 ft

h2 = 5 ft

h3 = 20 ft

g = 32.2 ft / s

Solution:

For Ball A:

aA = –gt

vA = –gt

sA = (–g/2)t^2 + h1

For Ball B:

aB = –g

vB = –gt + vB0

sB = (–g/2)t^2 + vB0t + h2

Guesses:

t = 1 s, vB0 = 2 ft/s

Given:

h3 = (–g/2)t^2 + h1

h3 = (–g/2)t^2 + vB0t + h2

t = 1.115 s ….. Answer

vB0 = 31.403 ft/2 ….. Answer

Now the question is:

FIRST

Please refer to "For Ball B" data, why the vB0 is existing as we can assume that the initial velocity of the Ball B is Zero?

SECOND

Why the value of t is 1 second and vB0 is 2 feet per second on "Guesses"?
Can I replace them with another Number?

THIRD

How to obtain the following Equation on "Given"?

h3 = (–g/2)t^2 + h1

These were very hard for me to understand, so, any response would be Highly appreciated.

Cheers

2. May 12, 2007

### Staff: Mentor

Ball B is thrown upward. Why would you think it's initial speed is zero?

I have no idea what "Guesses" mean. Are they expecting you to give a rough guess before you figure it out? Guess anything you want, as long as you then figure it out correctly.

They are just plugging the final position (where the balls pass each other) into your equation for Ball A (from your solution):
sA = h3