Recent content by akshajkadaveru

oh... i thought that't just the length of how far the thing went... how do i get that then >?

so is it... 1^2+1.44^2=whatever sqrt(whatever)= ______ @ tan^-1(1.44/1)? will that be my answer?

i didn't post anything on 27

ARGHGHGHH! so what i JUST did was wrong but what will my X and Y value things be then ?

2^2+3.46^2=15.97 sqrt(15.97) 3.99 m/s tan^-1(3.46/2)= 59.97 degrees 3.99m/s @ 59.97 degrees'

i know how to find the angle.. but how do i find the overall speed... is that by the pythagorean theorem and i plot that based on what i got for the V0x and v0y?

so for D... i already have the x y values and the angle right? so now i just have to plot them or something >

HOLY MOLY...in all honesty.. am i over complicating this ?

-1.44?

-9.8(.5)+v=3.46 v=8.36

oh...sorry accel = -9.8 initial = 3.46 time = .5 seconds

who's ur teacher ?

acceleration is 0 initial speed is 4m/s time .5 is that correct? after that, what do i need to do ?

i really appreciate you helping me but i don't understand how this is going to help me for the problem... a=dv/dt aDt=(v-v0) adT-v=-v0 -aDt+v=v0

dv/dt ?