Question

Free fall.

Answer 1

ok.. whats ur solution?

Answer 2

@ememlove
Even if you're stuck, can you draw a picture of the first part of the problem?

Answer 3

I get the feeling she's afk but still logged in...

Answer 4

Or all of this Tomfoolery has infected users as well :P

Answer 5

@AllTehMaffs join the physics chat? I'm on there.
I can't message you cause you don't reply.
& idk, what OS are you running on your computer?

Answer 6

hi @Mashy @AllTehMaffs @kittiwitti1 yeah i was afk just now sry, btw here i went abt with the first part, idk if correct or not. The second part im not sure how to do, would u guys be kind to help me? (:

Answer 7

Don't tag me, I know nothing about this o w o;;
Ask @Mashy

Answer 8

kitti.. u should be able to solve this with ease now !!

Answer 9

is my approach fine?

Answer 10

@Mashy You know I'd end up getting it wrong one way or another.
And I still have my own physics homework to finish... = n =;;

Answer 11

But I'll give it a try, I guess.
---
Firstly, I think we should use the kinematics equations.
Here is the given information for the first part:\[\Delta y = 200m\]\[t=1.0, 2.0, 3.0\]\[v_0=0m/s\]\[v_f= ?m/s\]
So we are required to find the final velocity for each of the given times. I will input these values into the following equation\[v_f=\cancel{v_ot}^0 -\frac{1}{2}at^2\]

Answer 12

you guys in college? haha, anw for 2nd part, i used \[\Delta x =v _{o}t-\frac{ g t^2 }{ 2 }\] to get the time it land to the ground @ -200m. then i used proportions to get my x's at respective time and then use these to plug in to v=x/t to get the respective velocities.

Answer 13

hmmm you have point there.. i should try tht for the 1st one

Answer 14

btw the eqn is abit off i guess \[v _{f}=v _{i}+at\]

Answer 15

if \[v _{f}^2=v _{i}^2+2a \Delta x\] you mean this?

Answer 16

Excuse my error... the equation we are using should be this one, because we don't have acceleration.\[\Delta y=\frac{1}{2}(\cancel{v_0}^0+v_f)t\]
For t=1.0:\[200=\frac{1}{2}(0+v_f)1.0\]\[400=1.0 \times v_f\]\[v_f=400m/s\]

Answer 17

we have aceleration due to gravity right? which is -9.8 m/s^2, coz the coin is being dropped from height

Answer 18

I'm in 11th grade, not college.
But I do get that a lot.
---
Well, yes... but we can also use this equation...
@Mashy am I right?

Answer 19

Anyway, do you have the answers to this question? You can check the results I got with that.

Answer 20

unfortunatey i dont have so i turn into this tool haha , trying my luck

Answer 21

Meh.
Well, until I can confirm that my methods are correct, I can't help you, sorry.

Answer 22

On second thought...\[v_f=\cancel{v_0t}^0+\frac{1}{2}gt^2\]
---
t=1.0 s...\[v_f=0+\frac{1}{2}(9.8m/s^2)(1.0)^2=4.9m/s\]---
t=2.0 s...\[v_f=0+\frac{1}{2}(9.8m/s^2)(2.0)^2=19.6m/s\]---
t=3.0 s...\[v_f=0+\frac{1}{2}(9.8m/s^2)(3.0)^2=44.1m/s\]

Answer 23

yup! i think using that formula is the most legit way haha then i used the vf^2 formula to get the velocity at that point in time right?

Answer 24

...I get the distinct feeling I did that wrong...

Answer 25

no i think its correct (: thanks!

Answer 26

this formula can be applied in part 2 where the inital velocity is 2m/s

Answer 27

SORRY! I did do it wrong. It's not Vf in the equation but the y-distance...

Answer 28

yeah it should be distance, looks like you're so used to write it like that

Answer 29

i also did not notice i thought it was distance haha

Answer 30

No, I just forgot the exact formula, it's been almost 2 weeks since we did kinematics.
Your way of using the formula was right:\[v_{fy}^2=v_{0y}^2+2g\Delta y\]

Answer 31

This might help - I need to finish my homework, sorry T^T
http://www.physicsclassroom.com/class/1dkin/u1l6a.cfm

Answer 32

does 11th grade study 2D projectile motions?

Answer 33

okay thanks for the help! (: gdluck!

Answer 34

If that was 2D, then yes.
So the equations I did gave you the distance, and you use the vf^2 equation to find the velocity.

Answer 35

Thanks for the medal, btw! (: