Can Any1 Help Me With A Simple Math Problem...?
master_wong
Join Date: 2004-11-05 Member: 32649Members
Join Date: 2004-11-05 Member: 32649Members
in Off-Topic
<div class="IPBDescription">gravity and time of falling...</div> ok, u can call me stupid, but my head really hurts today and i dont know how i forgot to do this simple question...
Ok, the acceleration of earth is 9.8m/s/s
*everything is ideal, so no air resistance is involved in this*
a printer/coin/whatever is dropped from top of the tower, which is 1000m of height.
how long will it take to drop down onto the ground?
plz show me the equation and some simple explaining.
thx!
Ok, the acceleration of earth is 9.8m/s/s
*everything is ideal, so no air resistance is involved in this*
a printer/coin/whatever is dropped from top of the tower, which is 1000m of height.
how long will it take to drop down onto the ground?
plz show me the equation and some simple explaining.
thx!
Comments
Seriously......WTH?
I hate math.
I hate math. <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
I concur.
/me crawls into a ball in the corner and cries
Seriously......WTH?
I hate math. <!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
<!--emo&???--><img src='http://www.unknownworlds.com/forums/html/emoticons/confused-fix.gif' border='0' style='vertical-align:middle' alt='confused-fix.gif' /><!--endemo-->
distance = (.5) * (acceleration) * (time) * (time)
* = "multiplied by"
distance = 1000m
acceleration = 9.8
solve for time
yay?
1000=4.9*x^2
204.08...=x^2
sqr(204.08)=sqr(x^2)
14.29...=x
TIME = 14.29... Sec
physics.
i have trouble remembering equations, but i think im pretty good at solving them.
physics.
i have trouble remembering equations, but i think im pretty good at solving them. <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
You had a formula and a calculator, man. There's not much to think about!
a = acceleration
okay great, what if I want speed?
take the integral
a*t + c
where c = initial velocity, in this casde zero (0)
so we have v = a*t
fairly obvious
but we want distance
take the integral again
.5*a*t^2 + c
where c is the initial distance, in this case 1000
a = -9.8
how long until the object reaches a height of 0?
0 = .5(-9.8)*t^2 + 1000
sqrt(2000/9.8) = t
Are you taking over for EEK or something?
Are you taking over for EEK or something? <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
Not sure... is that a bad thing?
you just stick in the values.
Its not lazy either, when you get to the way more advances stuff, you have to do
these simple calulations many many times, and thats way to hard to do by hand every
time.
Differentiating an equation gives you the rate of change of that equation, integrating is the reverse.
Velocity is rate of change of distance (think about it, it makes sense). Acceleration is rate of change of velocity. So differentiating velocity gives you acceleration, integrating acceleration gives you velocity, and integrating velocity gives you distance (not that that's all that helpful seeing as it gives you the <i>formula</i> for distance, which is simply "distance" itself because the formula for velocity uses distance.)