How Do You Calculate Vertical Motion and Acceleration?

AI Thread Summary
A ball is thrown upward at an initial speed of 80 ft/sec, and the discussion revolves around calculating its maximum height, speed after 3 seconds, and height at that time. The calculations provided initially resulted in incorrect values due to misapplication of formulas and sign errors. The correct approach emphasizes that gravitational acceleration should be treated as negative, affecting the velocity and height calculations. Clarifications were made regarding positive and negative heights and velocities, with the consensus that height is positive when above the starting point and velocity is negative when descending. Understanding these concepts is crucial for accurately solving projectile motion problems.
aldeemativo
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Homework Statement



A ball is thrown upward with an initial speed of 80 ft/sec.
Find the following:
a. how high does it go?
b. after 3 seconds, what is the speed of the ball?
c. how high is it at this time?


Homework Equations



a. how high does it go?
b. after 3 seconds, what is the speed of the ball?
c. how high is it at this time?


The Attempt at a Solution




i had come up from this solution:

given: Vo = 80 ft/sec
Vf = 0 ft/sec
where Vo=original velocity , Vf is the final velocity , t is the time , and h as the height

solution: a.) using formula: h = Vf2 - Vo2 over 2(32ft/sec2)
and as I substituted for the values I derived at my answer of -100 ft

solution: b.) using formula: Vf = (32ft/sec2)t + Vo
and as I substituted for the values I derived at my answer of 7,680ft/sec


solution: c.) using formula: h = Vot - 1/2 gt2
and again as I substituted for the values I derived at my answer of 96 ft


can someone correct this for me? this is my exam for tomorrow. thank u so much.
 
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aldeemativo said:

Homework Statement



A ball is thrown upward with an initial speed of 80 ft/sec.
Find the following:
a. how high does it go?
b. after 3 seconds, what is the speed of the ball?
c. how high is it at this time?


Homework Equations



a. how high does it go?
b. after 3 seconds, what is the speed of the ball?
c. how high is it at this time?


The Attempt at a Solution




i had come up from this solution:

given: Vo = 80 ft/sec
Vf = 0 ft/sec
where Vo=original velocity , Vf is the final velocity , t is the time , and h as the height

solution: a.) using formula: h = Vf2 - Vo2 over 2(32ft/sec2)
and as I substituted for the values I derived at my answer of -100 ft

solution: b.) using formula: Vf = (32ft/sec2)t + Vo
and as I substituted for the values I derived at my answer of 7,680ft/sec


solution: c.) using formula: h = Vot - 1/2 gt2
and again as I substituted for the values I derived at my answer of 96 ft


can someone correct this for me? this is my exam for tomorrow. thank u so much.
We can help you with homework questions, but not with exam questions. Is this a take home exam or just a sample of what might be asked? Actually, your equations are good but your plus/minus signs and math is not so good. It is sometimes best to leave off the units in the equation and put the corrrect units back in at the end.
 
PhanthomJay said:
We can help you with homework questions, but not with exam questions. Is this a take home exam or just a sample of what might be asked? Actually, your equations are good but your plus/minus signs and math is not so good. It is sometimes best to leave off the units in the equation and put the corrrect units back in at the end.

this is just an example problem. our exam would be similar as this...
 
Speed finder... help

Homework Statement



a ball is thrown vertically upward with an initial speed of 80 ft / sec.

Homework Equations



after 3 seconds, what is the speed of the ball?

The Attempt at a Solution



using the formula:Vf = Vo +gt
where Vf is the final velocity
Vo is the original velocity
g as the gravitational pull : 32ft/sec
and t as the time...

176 ft/ sec.. is this the right answer?
 


This is the formula you should use: v(t) = v0+at (Let v(t) be the velocity after t seconds.)
Since gravity pulls the ball down, and not up, the acceleration will have a negative value.

a = -g = -32 ft/s2

I hope that helped.
 


How did you get -100 ft?

Displacement can be negative, it just means it is in the negative direction with respect to the starting position (whether this makes sense in the context of a question is another story).

At the projectiles maximum height, what do you know about its velocity?
 


thanx. amm can i ask another thing? how about height??... when can we say that the displacement of a motion is negative if the forces discussed are those relevant to vertical direction... is it when it goes up or is it when it goes down that we can say it has negative height? thanx...
 
aldeemativo said:

Homework Statement



A ball is thrown upward with an initial speed of 80 ft/sec.
Find the following:
a. how high does it go?
b. after 3 seconds, what is the speed of the ball?
c. how high is it at this time?


Homework Equations



a. how high does it go?
b. after 3 seconds, what is the speed of the ball?
c. how high is it at this time?


The Attempt at a Solution




i had come up from this solution:

given: Vo = 80 ft/sec
Vf = 0 ft/sec
where Vo=original velocity , Vf is the final velocity , t is the time , and h as the height

solution: a.) using formula: h = Vf2 - Vo2 over 2(32ft/sec2)
and as I substituted for the values I derived at my answer of -100 ft
The value of g is negative, so your answer is numerically corect, but of course it reaches a height of + 100 feet.
solution: b.) using formula: Vf = (32ft/sec2)t + Vo
and as I substituted for the values I derived at my answer of 7,680ft/sec
Again g is negative, and t =3 and v0 is 80...your math is way off, how can it be traveling that fast ?? Solve for Vf, and which way is it moving at that time, up or down?
solution: c.) using formula: h = Vot - 1/2 gt2
and again as I substituted for the values I derived at my answer of 96 ft


can someone correct this for me? this is my exam for tomorrow. thank u so much.
Looks good!
 


It is normal to let the ground or the starting point be at 0 ft. Anywhere above this point, is a positive height, and anywhere below is negative.

If you mean the velocity, it is usually positive when the object is on its way up, and negative when the object is on its way down.

Note: If an object is above the ground, but on its way down, the height above the ground is positive, but the velocity is negative. That is an important difference.
 
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