*Need Help: Acceleration and gravity calculation - Thanks

AI Thread Summary
The discussion revolves around calculating acceleration due to gravity using the formula d = 1/2at², where d is the distance fallen and t is the time taken. The user provides specific values: a distance of 4.42 meters and two drop times of 0.707 seconds and 0.79 seconds. Participants clarify the formula and emphasize the need to substitute the values into the equation to solve for acceleration. There is also a suggestion that the user should attempt the calculation before seeking help. The conversation highlights the importance of correctly applying the formula to determine the acceleration due to gravity.
nukeman
Messages
651
Reaction score
0

Homework Statement



This is for a Astronomy class, and I have to go over a few examples. But I need help understand the proper steps in calculating this. I will give you as much information as you can.

I have this formula

"For an object dropped from rest, the distance of fall d is related to the time of fall t and acceleration due to gravity a by

d = 1/2at to the 3rd (the 1/2 is 1 over 2)

If we measure d and t we can solve for the acceleration, a:

a = 2d / t squared (2d / t squared mean over, not divided by)

Now my variables for the calculation are simple.

Distance: 4.42 meters

object drop 1: .707 seconds
objects drop 2: .79 seconds

Can anyone help me?





Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
nukeman said:

Homework Statement



This is for a Astronomy class, and I have to go over a few examples. But I need help understand the proper steps in calculating this. I will give you as much information as you can.

I have this formula

"For an object dropped from rest, the distance of fall d is related to the time of fall t and acceleration due to gravity a by

d = 1/2at to the 3rd you mean t squared[/color]
(the 1/2 is 1 over 2)

If we measure d and t we can solve for the acceleration, a:

a = 2d / t squared
yes
(2d / t squared mean over, not divided by)
same thing) example: 1 'over' 2 is the same as 1 divided by 2)
Now my variables for the calculation are simple.

Distance: 4.42 meters

object drop 1: .707 seconds
objects drop 2: .79 seconds

Can anyone help me?
Are these values determined in a lab experimemt, and you need to calculate 'a', the acceleration due to gravity? They don't look right...
 
your 1st post doesn't make a whole lot of sense if you read back over it. I think you mean d=1/2at^2.
you also don't say what your trying to find out, I assume its gravity(a)
also your meant to have an attempt at solving it before posting.
anyway - very simple put the numbers in your formula and solve for a.
 
whoops posted at the same time and virtually the same thing.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top