Recent content by kayte

alrighty. thank you so much for your help! :biggrin:

yes we are "As distance increases, what trend(s) do you see in the values for 'acceleration at end'?"

it's a sub question dealing with trends as distance increases

hmm... oh! i see haha :wink: one more question. what does the phrase "acceleration at end" mean and how can i find it?

ohh, like why did you tell me to use y=0? is that like the start/end of the path of the projectile?

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...