Calculating g with known time intervals

[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.

 

[cnh1995]

Welcome to PF!

Do you have a clearer image?

 

[dantechiesa]

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?

 

[cnh1995]

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?

 

[dantechiesa]

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.

 

[cnh1995]

Particularily vf.

You don't need Vf.

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?

 

[dantechiesa]

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?

 

[cnh1995]

= 2.489 m/s^2

Does this look right?

Is that the value of 'g' in England?

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).

 

[dantechiesa]

Is that the value of 'g' in England?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?

 

[cnh1995]

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.

 

[dantechiesa]

We can't derive the second equation as we don't know the distance from those points?

d = 0 - .5(.515)^2a

 

[cnh1995]

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.

7.69 = v0(1.67) + .5a(1.67)^2

Or just modify this equation using the correct value for t.

 

[dantechiesa]

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?

 

[cnh1995]

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.

 

[dantechiesa]

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)

 

[cnh1995]

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?