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

  • Thread starter kayte
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  • #1
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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.)
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi kayte! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
Vx (would this be horizontal vector?) = d/t = 5 m/1 s = 5 m/s
yup! :smile:
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!! :wink:

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 :smile:
 
  • #3
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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 of the path to find ∆y?
 
  • #4
tiny-tim
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hi kayte! :smile:

(just got up :zzz:)

you're confused between v constant and a constant …

a is -g :wink:

(and if A and B are at the same height, then t = 1 isn't at y = maximum, is it?)
 
  • #5
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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
 
  • #6
tiny-tim
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hmm … 2 variables, only 1 equation :frown:

try using ∆t = 1, ∆y = 0 o:)
 
  • #7
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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?
 
  • #8
tiny-tim
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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 :smile:

now you can find ∆y for t = 1/2 ! :wink:
this will be the vectors at catch point …
i don't understand :confused:
 
  • #9
9
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ohh, like why did you tell me to use y=0? is that like the start/end of the path of the projectile?
 
  • #10
tiny-tim
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because you know both y and t at that point …

you didn't know both y and t at the top point :wink:
 
  • #11
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hmm... oh! i see

haha :wink:

one more question. what does the phrase "acceleration at end" mean and how can i find it?
 
  • #12
tiny-tim
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one more question. what does the phrase "acceleration at end" mean and how can i find it?
where does it come from? :confused:

(it's not in the question, and every projectile has the same acceleration anyway)
 
  • #13
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it's a sub question dealing with trends as distance increases
 
  • #14
tiny-tim
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are we still talking about projectiles?

can you give a quote? :confused:
 
  • #15
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yes we are

"As distance increases, what trend(s) do you see in the values for 'acceleration at end'?"
 
  • #16
tiny-tim
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the acceleration is constant :confused:
 
  • #17
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alrighty. thank you so much for your help! :biggrin:
 

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