alrighty, so...
Δy=vinitialΔt+(1/2)a(Δt)2
0=vinitial(1)+(1/2)(-9.81)(1)2
0=vinitial-4.905
4.905=vinitial
this will be the vectors at catch point because time is 1 and height is 0?
oh O:
so...
Δt=.5 s (because the max height is in the middle)
a=-9.81 m/s2
Δy=vinitialΔt+(1/2)a(Δt)2
∆y=vinitial(.5)+(1/2)(-9.81)(.5)2
∆y=(.5)vinitial+1.22625
okay... so i chose to use the equation Δy=vinitialΔt+(1/2)a(Δt)2
knowing the given:
Δt=1 s
a=0 m/s2 (because it's constant)
i substitute:
Δy=vinitialΔt+(1/2)a(Δt)2
∆y=vinitial(1)+(1/2)(0)(1)2
∆y=vinitial+0
∆y=vinitial
and I'm stuck here... :frown: would i need to know max height...
Homework Statement
I need to find the horizontal vectors and vertical vectors at launch, maximum height and at final catch of a projectile. I know that the distance between point A and point B is 5 and the total time is 1. Homework Equations
How can I also find the maximum height and the...