Recent content by poh

I see now Thank you :D

it's take same amount of time to go up as go to down. -1.1s is from ground level so when 1.9s-1.1s its where they meet and point is 0 and the diffrence is time from the top of the arc to ground,

Vi = 4m/s I think it was all the time vertical velocity

ah I see now so Vf=Vi+at where Vf=0 Vi 4m/s and a=9.8 m/s t=? so t= (Vf-Vi)/a t= -4m/s / -9.8 m/s^2 t= .41s from top to the ground or I should 1.9 s- 1.1s = .8s and that's time from top to ground ?

I was asked how longs takes to reach the ground. so answer is 1,9s How long till it reaches the top of its arc? is Vf=Vi + at where a=0 and t=1.9s ?

Negative means that's going down?

-1.1 and 1.9s

I did mistake in calculations.. my new calculations are 1.8 or 1.1

i got 1.1s

divede t by itself or I should use square roof to take out t^2 ?When I assume Viy = 0 then I got 1.4s

I got t= 2.9s

from 10 m tall tree to the ground

t is time how much time takes to fall ?

acceleration should be -9.8m/s^2

for b) i tried Yf= yi + Viyt + 1/2 at^2 0=10m +4m/s*t + 1/2(9.8m/s)t^2 t= 2*(10m+4m/s)/9.8 m/s t= 28/9.8s t= 2.9s t is time to reaches the ground from the top Or i should assume my Yi is 0?