Calculating g with known time intervals

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Homework Help Overview

The discussion revolves around calculating the acceleration due to gravity (g) using known time intervals and kinematic equations. The original poster is attempting to analyze the motion of a ball thrown vertically, focusing on the time it takes to reach a certain height and the relationship between initial velocity, acceleration, and time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to relate initial velocity, acceleration, and time. There are attempts to derive relationships between these variables while questioning the need for final velocity (vf) and the implications of time intervals on the calculations.

Discussion Status

Participants are actively engaging in the problem, offering guidance on how to set up equations and questioning assumptions about the motion. There is a mix of interpretations regarding the time intervals and the necessary equations to use, with some participants suggesting modifications to existing equations based on the given vertical displacement.

Contextual Notes

There are constraints regarding the clarity of the provided image and the specific values needed for calculations, which are causing some confusion among participants. The discussion also highlights the importance of understanding the relationship between time taken to reach the top of the trajectory and the subsequent descent.

dantechiesa

Homework Statement


In the photo[/B]
https://imgur.com/a/tf4jI

Homework Equations


vf^2 = v0^2 + 2ad
d = v0t + 1/2 a t^2
3. The Attempt at a Solution [/B]
Soo, I am really having difficult on where to start, however, I believe I need to utilize the fact that at the top of the path, the v0 is zero. And utilizing that and the difference in time between the intervals to calculate the a. However, without vf I am not sure how to go about.

Can someone lead me in the correct direction.

Thanks.
 
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Welcome to PF!

Do you have a clearer image?
KlbqGrc_d.jpg
 
cnh1995 said:
Welcome to PF!

Do you have a clearer image?
View attachment 211906
Hmm, I am not sure why its so unclear for you. Did you click on the imgur link directly?
 
dantechiesa said:
Hmm, I am not sure why its so unclear for you. Did you click on the imgur link directly?
Yes. But I am able to read it, just have to zoom it a little.
Consider the motion in y-direction only.

The ball is thrown vertically with some velocity v0. It falls back to the same level after 2.70s. How would you write this as a kinematic equation?
 
cnh1995 said:
Yes. But I am able to read it, just have to zoom it a little.
Consider the motion in y-direction only.

The ball is thrown vertically with some velocity v0. It falls back to the same level after 2.70s. How would you write this as a kinematic equation?
d = 0
a = ?
v0 = ?
t = 2.70

So the equation would be :
0 = v0 (2.70) + .5 a (2.70)^2

But I just can't figure out how to break this down with two missing variables. Particularily vf.
 
dantechiesa said:
Particularily vf.
You don't need Vf.
dantechiesa said:
So the equation would be :
0 = v0 (2.70) + .5 a (2.70)^2
Right. So what is the relation between v0 and g from this equation?

Can you write one more similar equation using the given vertical distance between the two levels?
 
cnh1995 said:
You don't need Vf.

Right. So what is the relation between v0 and g from this equation?

Can you write one more similar equation using the given vertical distance between the two levels?

Rearranging the initial equation gives the relationship of
v0 = 1.35a

and I can create another equation between the two levels:
7.69 = v0(1.67) + .5a(1.67)^2

wait. do I then just sub in the v0 = 1.35 into the second equation.

7.69 = 1.35(1.67)a + .5(1.67)^2a

7.69 = a(1.35*1.67 + .5*1.67^2)
7.69 / (1.35*1.67 + .5*1.67^2) = a

a = 2.107 m/s^2

Does this look right?
 
dantechiesa said:
= 2.489 m/s^2

Does this look right?
Is that the value of 'g' in England?:wink:

dantechiesa said:
and I can create another equation between the two levels:
7.69 = v0(1.67) + .5a(1.67)^2
No. How much time will have elapsed when the ball gets at a height of 7.69m from the starting point and is coming downward?
(You already have the total time of its journey).
 
cnh1995 said:
Is that the value of 'g' in England?:wink:No. How much time will have elapsed when the ball gets at a height of 7.69m from the starting point and is coming downward?
(You already have the total time of its journey).
Sorry! That just confused me a bit. The ball travels 7.69m in 1.67 seconds. The time from the top point, where v = 0 and the tip of 7.69 m is .515s?
 
  • #10
dantechiesa said:
The time from the top point, where v = 0 and the tip of 7.69 m is .515s?
Yes. So what is the second equation?

And in both your equations, the sign of the acceleration should be minus, since initially it is acting against the given velocity.
 
  • #11
We can't derive the second equation as we don't know the distance from those points?

d = 0 - .5(.515)^2a
 
  • #12
dantechiesa said:
We can't derive the second equation as we don't know the distance from those points?

d = 0 - .5(.515)^2a
You know the vertical displacement of the ball from the starting point, you know the time elapsed.
Write a similar equation using these values.
dantechiesa said:
7.69 = v0(1.67) + .5a(1.67)^2
Or just modify this equation using the correct value for t.
 
  • #13
cnh1995 said:
You know the vertical displacement of the ball from the starting point, you know the time elapsed.
Write a similar equation using these values.

Or just modify this equation using the correct value for t.
Im sorry, I don't comprehend, I am having a really difficult time wrapping my head around the idea.

would t be .835s?
 
  • #14
dantechiesa said:
Im sorry, I don't comprehend, I am having a really difficult time wrapping my head around the idea.

would t be .835s?
No.
How much time does the ball take to reach the topmost point? How much time does it take to fall from the topmost point to a height of 7.69m from the starting point?
The additiom of these two will be the value of t in your above equation.
 
  • #15
cnh1995 said:
No.
How much time does the ball take to reach the topmost point? How much time does it take to fall from the topmost point to a height of 7.69m from the starting point?
The additiom of these two will be the value of t in your above equation.
t = 1.865. (taking the halves of the two times and adding them)
The time to reach the top of the H is .515 seconds. (half of 1.03)
 
  • #16
dantechiesa said:
t = 1.865. (taking the halves of the two times and adding them)
The time to reach the top of the H is .515 seconds. (half of 1.03)
Correct. So what value did you get for g?
 

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