Here's the question you clicked on:
stracyl
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
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
hmm i think something wrong in there.. @iNeedTheHelp can you tell which formula you have used?
if i am right u have used s=ut+0.5at^2.. right?
Yeah is there a flaw in the equation?
well if you have use this formula , in it u=initial velocity. and in question initial velocity is given.
Yes I already know that but what is wrong with this equation and sorry I'm kind of tired
its ok.. well your equation will be then s = 24t+0.5at^2
you missed 24t in your equation.
ohhh wow thats a big one too sorry but @nubeer you can help him because I am going to bed
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