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

 

[m2003]

Hello,

Kinetics is a branch of physics that deals with the study of motion and forces acting on objects. It is a complex subject and can be challenging to understand, so it is normal to have difficulties with problems like the one you mentioned.

To answer your questions:

FIRST: The initial velocity of ball B is not necessarily zero because it is thrown upward from the ground. This means that it already has some initial velocity when it passes ball A at a height of 20 ft. Therefore, we need to consider the initial velocity, vB0, in our calculations.

SECOND: The values of 1 second and 2 feet per second for t and vB0 were just guesses made by the person who provided the solution. They are not the only possible values and you can definitely replace them with other numbers. However, in order to find the correct value for t and vB0, you would need to solve the equations for ball A and B simultaneously using the given information.

THIRD: The equation for "Given" was obtained by setting the position of ball A and B to be equal at a height of 20 ft. This gives us the equation: (–g/2)t^2 + h1 = (–g/2)t^2 + vB0t + h2. We can then solve for t and vB0 using this equation.

I hope this helps clarify things for you. Kinetics is a challenging subject, so don't be discouraged if you are struggling with it. Keep practicing and seeking help when needed. Good luck!