How Is the Initial Velocity of Ball B Calculated in Kinetics Problems?

[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.

ANSWER

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

 

[Doc Al]

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?

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

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.

THIRD

How to obtain the following Equation on "Given"?

h3 = (–g/2)t^2 + h1

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

For Ball A:

aA = –gt

vA = –gt

sA = (–g/2)t^2 + h1

sA = h3