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

[kayte]

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.)

 

[tiny-tim]

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 v_initial, that's what you're trying to find!

you have a and t and (y - y_initial) = 0, and you want v_initial

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

show us what you get ​

 

[kayte]

okay... so i chose to use the equation Δy=v_initialΔt+(1/2)a(Δt)²

knowing the given:
Δt=1 s
a=0 m/s² (because it's constant)

i substitute:
Δy=v_initialΔt+(1/2)a(Δt)²
∆y=v_initial(1)+(1/2)(0)(1)²
∆y=v_initial+0
∆y=v_initial

and I'm stuck here... would i need to know max height of the path to find ∆y?

 

[tiny-tim]

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?)

 

[kayte]

oh O:

so...
Δt=.5 s (because the max height is in the middle)
a=-9.81 m/s²

Δy=v_initialΔt+(1/2)a(Δt)²
∆y=v_initial(.5)+(1/2)(-9.81)(.5)²
∆y=(.5)v_initial+1.22625

 

[tiny-tim]

hmm … 2 variables, only 1 equation

try using ∆t = 1, ∆y = 0 ​

 

[kayte]

alrighty, so...

Δy=v_initialΔt+(1/2)a(Δt)²
0=v_initial(1)+(1/2)(-9.81)(1)²
0=v_initial-4.905
4.905=v_initial

this will be the vectors at catch point because time is 1 and height is 0?

 

[tiny-tim]

alrighty, so...

Δy=v_initialΔt+(1/2)a(Δt)²
0=v_initial(1)+(1/2)(-9.81)(1)²
0=v_initial-4.905
4.905=v_initial

looks ok

now you can find ∆y for t = 1/2 !

this will be the vectors at catch point …

i don't understand

 

[kayte]

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]

because you know both y and t at that point …

you didn't know both y and t at the top point ​

 

[kayte]

hmm... oh! i see

haha

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

 

[tiny-tim]

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)

 

[kayte]

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

 

[tiny-tim]

are we still talking about projectiles?

can you give a quote?

 

[kayte]

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

 

[kayte]

alrighty. thank you so much for your help!