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- anonymous

an object is thrown directly up with a velocity of 20.0 m/s and a d0=0. Determine how long it takes to get to the maximum height of 24.0 m

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- anonymous

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- anonymous

Well we know the maximum height occurs when the velocity is at 0
(initial velocity)(time) = (0.5)a(t^2)
Diviide both sides by 't
velocity = at/2
time = 2v/a = (2)(20.0) / (9.81 m/s/s)
time = 4.077sec

- nubeer

hmm i think something wrong in there.. @iNeedTheHelp can you tell which formula you have used?

- anonymous

let me see hmm

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- nubeer

if i am right u have used
s=ut+0.5at^2.. right?

- anonymous

Yeah is there a flaw in the equation?

- nubeer

well if you have use this formula , in it
u=initial velocity.
and in question initial velocity is given.

- anonymous

Yes I already know that but what is wrong with this equation and sorry I'm kind of tired

- nubeer

its ok.. well your equation will be then
s = 24t+0.5at^2

- nubeer

you missed 24t in your equation.

- anonymous

ohhh wow thats a big one too sorry but @nubeer you can help him because I am going to bed

- nubeer

sure no problem :) have sweet dreams
ok @stracyl
formula is this
s=ut+0.5at^2..
s=24, u = 20 a=9.81
24 = 20t+0.5(9.81)t^2
0.5(9.81)t^2 +20t -24 = 0
use quadratic formula and find the roots of t

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