I need help on KINETICS

[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

 

[pmonon]

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

Ans. for first: Ball b is thrown against g and to have a motion it must have a non zero initial velocity.

second: I didnt understand the thing with guesses. Why you need to guess?

third: the ball is already at a height h1 and it is moving against the direction of height.

 

[marko_benigno]

There are no guesses in this problem. Basically, there are two instances to solve the unknown.... one is when is the initial point of reference wherein time t =0, and the other is the time when the two balls are at the same height which we need to compute. At this point, t = unknown.

Usinfg the equation for ball A, we could solve this since we know all the initial condition for A. So,

to solve for the the time when the dist travelled by ball A is 20 w/ ref to ground:

20 = -½gt² + 40, xa ; 20 = (-1/2)gt^2 ; 40 = gt²
t = 1.114556425s

So now, we could use this info to get the initial velocity of ball B. Plugging in the values: using the formula, x = xb + Vb*t+ -½gt², wherein xb is the initial position of ball b and Vb is the initial velocity of ball B.

At time t = 1.114556425s, x = 20ft from the given.

20 = 5 + Vb*t + -½gt², where -½gt² = 20 from the initial computation.

we get,
20 = 5 - 20 + Vb*t
35 = Vb*t
Vb = 35/1.114556425 = 31.402627 ft/s!

So, there are no guesses here.

going back to your other questions,
If it is stated that the ball is thrown upward or downward, it means that the ball has some initial velicty. If its released, thats the point we could say that its free fall meaning, the initial velicity is zero.

for the last question: How to obtain the following Equation on "Given"? h3 = –½gt² + h1 ,

this was derived for the ball A from the satndard equation,

x = xa + Va*t + ½at².... where xa = init position = 40;Va = init velocity = 0 ( dropped!); a= acceleration = -g; and final position = 20

so it was simplified to
h3 = –½gt² + h1

Hope this helps.