Question

I messed up. I was looking at the wrong problem on my homework, i still need help.. Lol

Answer 1

@SapphireMoon

Answer 2

the answer i put on the last question was right, but it was for the question before that one. no wonder it suddenly made sense... i was looking at the wrong problem...

Answer 3

i still need help... this is really awkward. I'm not smart lol

Answer 4

Totally fine. I felt like that when I first started kinematics.

Answer 5

If a ball is thrown off of a 123 m cliff at a speed of 18 m/s, how far from the base of the cliff is the ball?
on this one with the GUESS method

Answer 6

This is when it lands, yes?

Answer 7

i have no idea.. after reading it so many times, the question doesn't even make sense anymore.

Answer 8

Let's pretend it's when it lands XD otherwise there's all kinds of problems... Mkay, so first we want to list the givens. This is 2D kinematics, so we need a set for the x and the y. We're going to go from a frame of reference where up is the +y direction, right is the +x direction, and the ball is thrown to the right. 0 for both will be the ground.

\[x_{0} = 0m\]
\[x = ?\]
\[v_{0x} = 18m/s\]
\[v_{x} = ?\]
\[a = 0m/s^2\]
\[t = ?\]

So I rolled the givens and unknowns into one. The acceleration is 0 because in projectile motion there's always no acceleration in the x-direction. Can you do this yourself for the y?

Answer 9

wait isn't accelaration always 9.8 m/s?

Answer 10

or is that only sometimes?

Answer 11

and wouldn't the first x be 123m? or where would that go... i'm so confused..

Answer 12

Horizontal
V = x/t
Vertical
x = vit + ½ at2
vf = vi +at

these are the equations i can use.

Answer 13

No, the first x is 0 because we defined 0 as the building. 123 is going to be y-naught (the original height). a is 9.8 (or -9.8 in our case with up as +y) in the y-direction only. Can you try setting up the givens now? Or do you need me to do it?

Answer 14

can you do it, please.. I'll try to do the unknowns by myself

Answer 15

Well, I do both at once, but all right...

\[y_{0} = 123m\]
\[y = 0m\]
\[v_{0y} = 0m/s\]
\[v_{y} = ?\]
\[g = -9.8m/s^2\]
\[t = ?\]

So now we know what we have, and we know what we don't know. What equation do you think will help us?

Answer 16

i have no idea... honestly, you're doing a great job helping me, it's just none of it's clicking. i'm so lost and i don't even know the right questions to ask before i can get where i need to be.

Answer 17

like, i have no idea what y0 stands for, or g, or y, or vy, or v0y. i know that t= time, but that's it and that's the most basic thing...

Answer 18

i'm really sorry for wasting your time, but i have to take my brother to karate. thank you so much for trying to help me.

Answer 19

Sure, no prob, we can work this out when you get back? y0 is the initial height (the naught/0 always means initial) y is the final, g is acceleration by gravity, vy is final velocity in the y-direction, and v0y is initial velocity in the y direction.

Answer 20

You MAY find
http://perendis.webs.com
some help ...
hope so.

Answer 21

@faiithhh the lunar gemstone is helping you I see, which is good. To you, try not to get upset. Give yourself the time to time things. If you stay as relaxed and calm as you can, then I think you have a much better chance of getting things to click. If I could show you the mess I'm making of a problem at the moment, you'd probably be in stitches.
Yes, it is difficult. Einstein, Newton et al all found it difficult.
As President Kennedy may have said "... because it's hard ... "
good on you.