Vi for Constant Acceleration Kinematics

In summary, The conversation discusses using a formula to find the height of a flag using the time it takes a ball to pass the flag twice. The speaker has some questions about the concept and solving the equation, including why initial velocity is needed and why it cannot be zero. They also mention using a quadratic equation and the relationship between distance and time in motion.
  • #1
diego1404
6
0
So, there's a question that's 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 let's say t=5seconds and t=15 seconds

I know how to solve it but i don't 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 can't 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 don't 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 don't understand
mathematically you can plug in zero but why doesn't it make sense here>
 
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  • #2
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
.........
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
diego1404 said:
So, there's a question that's 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 let's say t=5seconds and t=15 seconds

I know how to solve it but i don't 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 can't 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 don't 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 don't understand
mathematically you can plug in zero but why doesn't it make sense here>

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.
 

What is Vi for Constant Acceleration Kinematics?

Vi for Constant Acceleration Kinematics refers to the initial velocity of an object undergoing constant acceleration. It is the velocity at the beginning of a motion, before any acceleration has taken place.

How is Vi calculated for Constant Acceleration Kinematics?

Vi can be calculated using the equation Vi = Vf - at, where Vf is the final velocity, a is the acceleration, and t is the time interval. Alternatively, Vi can also be calculated using the equation Vi = (Vf + Vo)/2, where Vo is the initial velocity.

What are the units for Vi in Constant Acceleration Kinematics?

The units for Vi in Constant Acceleration Kinematics are typically meters per second (m/s) or feet per second (ft/s), depending on the unit system being used.

Is the value of Vi always positive in Constant Acceleration Kinematics?

No, the value of Vi can be positive, negative, or zero in Constant Acceleration Kinematics. It depends on the direction of the initial velocity and the direction of the acceleration.

Why is Vi important in Constant Acceleration Kinematics?

Vi is important because it helps determine the motion of an object undergoing constant acceleration. It is a crucial component of the equations used to calculate displacement, final velocity, and time in this type of motion.

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