# Vi for Constant Acceleration Kinematics

1. Sep 20, 2012

### diego1404

So, theres a question thats been bugging me.
If a ball is thrown upward to find out the height of a flag by timing when the ball passes the flag how do you find the height given that the ball passes the flag twice. When only time is given. So lets say t=5seconds and t=15 seconds

I know how to solve it but i dont understand the concept
I will use  to ask questions along the way

What i would do is use the formula
x=xi +vi + (1/2)(-g)(t^2)

lets say Equation 1 is

x=vi(5)-4/9(.5^2)

then subtract t2-t1 and divide it by 2 so: (t2-t1)/2
this would give us time 2.5
we can get velocity when time is 2.5s by vf-va=a*t so, 0-va=-9.81(2.5)
Va=24.525

$#why cant i go initial time t=5 seconds we can get velocity by a*t =vf-vi where vi=0 Vf-vi= -49.05, vi=49.05 then substitute x with equation 1 earlier and va=24.25 then find out vi (va^2)-(vi^2)= 2gx$#why is Va(final velocity) not zero here if Vf(final velocity) was zero earlier when time was 2.5 seconds

(24.525^2)-(vi^2)=(2)(9.81)((vi(5)-4/9(.5^2))

then you use quadratic equation to find vi
plug in vi to the initial equation to get distance

 another question is why is initial velocity needed to find out the distance
 why is vi not zero since it starts at the bottom
 if it an object was dropped above a certain height its vi is zero
 what i was thinking as my main solution was that x=(-1/2)(9.81)(.5^2)
$# why was that wrong what i dont understand is that if this motion was horizontal ie. a wheel moving in the ground you can have a vi=0 but in the question above vi cannot equal zero. i always hear that distance is a function of time, which i dont understand mathematically you can plug in zero but why doesnt it make sense here> 2. Sep 20, 2012 ### azizlwl another question is why is initial velocity needed to find out the distance why is vi not zero since it starts at the bottom if it an object was dropped above a certain height its vi is zero what i was thinking as my main solution was that x=(-1/2)(9.81)(.5^2)$# why was that wrong
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Nothing wrong with equation but you got negative value. It works if it is well.

What you have found is V2i which is the velocity when it reached the top of the pole.
Since acceleration is constant, you need not find the initial velocity at the ground to find the height.
Sketch a velocity vs. time graph to give you the clue for the answer.

3. Sep 20, 2012

### PeterO

Addressing this last part.

What did you want to "plug in zero" to?

Suppose you decided to study traffic on a freeway - and by your choice, time zero was 12:00 noon.
You decision does not mean that all the cars had to park on the freeway and wait until 12:00 noon before they all start moving.

Also, if you decided that zero position was directly under the bridge you were observing from, it doesn't mean the cars have to line up 200 abreast under the bridge, waiting for 12:00 noon to arrive.