Hi everyone, I have a difficulty of the following Dynamics (kinetics) problems (from Hibbeler’s Eng Mech Dynamics, 11E book)(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Please help me on KINETICS

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