Recent content by Ahmedstein

Just don't put the files in the system folders of the iPod, put them in the main drive or in a separate folder that you create.

I think this problem is a bit advanced.Anyway, Thanks for the help everyone. You guys are great.

please read it again and tell me what that (t) represents the (t) in the equation (h = 0.5 g t^2) is the objects time it's not the same as the (t) in the problem

but dude, h is required in the problem, if I use this the answer is going to be: h = V (t - square root of 2h/g) that doesn't make any sense

yeah dude, my bad.. but it's not the average velocity it's just the x component of V which is constant at each point. anyway Y = 20*1.44 - 0.5*9.8*1.44^2 = 18.64 Hooooray

Does anyone have anything to say?

aaaahhhhhhh

:rolleyes: :frown: :cry:

shouldn't it be y(t) = V0 - 0.5 g t^2 , where V0 is the initial velocity, which in this case is equal to zero

maaan, you guys reply so fast!

you need to find the time when the horizontal displacement is 50, and then find the vertical displacement at that point. I got t= 5/3

so it is h= v (t- ?) :(

Ok, here it is. An object was dropped from the top of a cliff and the dropper heard the sound after (t) time. ok, and it's given in the question that speed of light is V, and free fall accerelation is a, and I should find the height of the cliff, so I figured out that I should use h= V * t, but...

actually, it does. It is a computed homework system and if I use variables that are not in the problem it says "the answer doesn't not depend on that variable" for example I tried that: h is height or displacement h= -0.5gt(squared) t= square root of (2h/g) I know it's a square root of...

the force of gravity?! :cry: