Question

I can't figure this out. I am trying to get a to be close to acceleration due to gravity which is 9.8. If you could help me I would greatly appreciate it.
.01+2.327(.05)+(1/2)a(.05^2). I think I am solving it wrong algebraically, If you could give me the steps I would like to know where I am making my mistake. Thank you

Answer 1

We'll need an equation at some point. "= ??"

Answer 2

I have that y(t)=y(initial)+v(initial)t+(1/2)at^2
My data is y=.01, v(initial)=2.327, t=.05, I need to calculate a which is the accelration

Answer 3

Ah! So the answer to my question is "= 0.01".

Okay, you'll have to make up your mind on a few things.

1) What is 0.01. Is that the result, y(0.05) = 0.01, or is that the initial displacement, \(H_{0} = 0.01\)?

2) Are you SURE the sign on your squared term should be the same as the sign on your linear term?

Answer 4

I have no idea I did this experiment in class I think y is our distance in that time interval. Heres what we did, we dropped a cylinder and calculated its velocity and time at a specific distance. The first distance was .01m and it took .05 seconds to get to .01m. the computer gave us the initial velocity

Answer 5

Yes i am sure of the terms

Answer 6

Doesn't sound right.

1) If you dropped it, the initial velocity was zero (0). Did you throw it?
2) You DO have to decide on a coordinate system. Which way is positive? Up or Down?

I'm thinking we have this:

Standard Formula: \(H(t) = H_{0} + t\cdot V_{0} + (1/2)at^{2}\)

1) Initial Height - Let's consider the launching point, \(H_{0} = 0\) and any displacement will be in the negative direction. Thus, DOWN is negative.

Now we have this:

\(H(0) = 0\)

\(H(t) = t\cdot V_{0} + (1/2)at^{2}\)

2) Initial Velocity. Since we dropped it, and we didn't throw it or otherwise launch it, \(V_{0} = 0\)

Now we have this:

\(H(t) = (1/2)at^{2}\)

3) Finally, we know that \(H(0.05) = -0.1 = (1/2)a(0.05)^{2}\).

This gives a = -8 -- That's actually closer than I thought, given that I completely ignored the value 2.327. I still don't know what that means. If it's the initial velocity, then we didn't drop it.

Answer 7

I was thinking the same thing that the initial velocity was zero. But the computer gave us an initial and final velocity. But I still do not get close to g or your answer. How do I solve for a, I think that is my problem. If I can get 8 that is close enough for me at this point

Answer 8

Let's work on that final velocity theory.

\(V(t) = a\cdot t\)

\(V(0.05) = -2.327 = a\cdot (0.05)\) and this produces \(a = -46.54\). Well, okay, that wasn't very pretty.

Well then, I'm officially confused. I showed you the development of a = -8, but that required simply ignoring one given value. Recognizing that value leads to a ludicrous value if we are on Earth. (It might be closer on Saturn!)

Good luck to you. We will need to understand the problem better.