# How can I find Horizontal and Vertical velocity, given distance and time?

## 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 acceleration?

## The Attempt at a Solution

Vx (would this be horizontal vector?) = d/t = 5 m/1 s = 5 m/s
Vy (would this be vertical vector?) = Vinitial + aΔt = (and i got stuck here because i'm not sure what v initial is equal to.)

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tiny-tim
Homework Helper
welcome to pf!

hi kayte! welcome to pf! (try using the X2 button just above the Reply box )
Vx (would this be horizontal vector?) = d/t = 5 m/1 s = 5 m/s
yup! Vy (would this be vertical vector?) = Vinitial + aΔt = (and i got stuck here because i'm not sure what v initial is equal to.)
you don't need vinitial, that's what you're trying to find!! you have a and t and (y - yinitial) = 0, and you want vinitial

so choose one of the standard constant acceleration equations that uses a t y and v …

show us what you get 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... would i need to know max height of the path to find ∆y?

tiny-tim
Homework Helper
hi kayte! (just got up :zzz:)

you're confused between v constant and a constant …

a is -g (and if A and B are at the same height, then t = 1 isn't at y = maximum, is it?)

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

tiny-tim
Homework Helper
hmm … 2 variables, only 1 equation try using ∆t = 1, ∆y = 0 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?

tiny-tim
Homework Helper
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
looks ok now you can find ∆y for t = 1/2 ! this will be the vectors at catch point …
i don't understand ohh, like why did you tell me to use y=0? is that like the start/end of the path of the projectile?

tiny-tim
Homework Helper
because you know both y and t at that point …

you didn't know both y and t at the top point hmm... oh! i see

haha one more question. what does the phrase "acceleration at end" mean and how can i find it?

tiny-tim
Homework Helper
one more question. what does the phrase "acceleration at end" mean and how can i find it?
where does it come from? (it's not in the question, and every projectile has the same acceleration anyway)

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

tiny-tim
Homework Helper
are we still talking about projectiles?

can you give a quote? yes we are

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

tiny-tim
the acceleration is constant alrighty. thank you so much for your help! 