Question

Where does the formula vy = .5gt come from?

Answer 1

\[{v_y} = \frac{1}{2}gt\]

Answer 2

It's generally derived with calculus - want to have a go at it? ^_^

Answer 3

I'm in AP Calc AB, so I'm down...

Answer 4

It'll be cake then ^_^ no lie.

So, you start with your definition of acceleration,
\[a=\frac{\mathrm{d}v}{\mathrm{d}t}\]
that multiply by dt and you get
\[a \mathrm{d} t =\mathrm{d}v\]
So we make that the indefinite integral
\[\int a \mathrm{d} t =\int\mathrm{d}v\]
What does that give you?

Answer 5

a is a constant in this scenario, btw, so yo can say

\[a \int \mathrm{d} t =\int\mathrm{d}v\]

Answer 6

Let's also actually just put boundaries on it, to make the constant behave. Time starts at zero, and your velocity starts at some value v0
\[a\int_0^t \mathrm{d} t =\int_{v_0}^v\mathrm{d}v\]

Answer 7

How would you evaluate
\[\int_0^x 1 \mathrm{d}x\]?

Answer 8

∫dx = x right?

Answer 9

Exactly! It's just x (plus a constant) evaluated from x to zero, right?
\[\int_0^x 1 \mathrm{d}x = (x+C)\big|_0^x\]
\[ = (x+C)-(0+C)=x\]
\[\int_0^x 1 \mathrm{d}x = x\]


Similarly, looking at the left hand side of the original equation
\[a\int_0^t \mathrm{d}t = a \Big(t+C\Big) \Big|_0^t=a(t)=at\]
\[a\int_0^t \mathrm{d}t=at\]

Now looking at the right hand side
\[ \int_{v_0}^v \mathrm{d}v = (v+C)\Big|_{v_0}^v=(v+C)-(v_0-C)=v-v_0\]
\[\int_{v_0}^v \mathrm{d}v=v-v_0\]
together they give the equation for the velocity of a particle given its acceleration and time


\[a\int_0^t \mathrm{d}t=\int_{v_0}^v \mathrm{d}v\]
\[at=v-v_0\]
\[v=at+v_0\]

magic!

Answer 10

Thanks, my Physics Teacher is going to be amazed!

Answer 11

^_^ Can you see where the next step comes is? this is just for the velocity, not the position.

Do you know the definition formula for velocity (like the one for acceleration above?)

Answer 12

v= d/t

Answer 13

Correct - do you know the differential form? (and if not, the form with Deltas?)

Answer 14

like,
\[a=\frac{\mathrm{d}v}{\mathrm{d}t}\]

whereas
\[v=?\]

Answer 15

\[a = \frac{ dd }{ dt }\]

Answer 16

Close! ^_^ To prevent having to write dd (which might be confusing) we'll use dx - just some change in x (or it could be in the y axis or the z axis or whatever - it's just a change in **position**)

So we have
\[v=\frac{\mathrm{d}x}{\mathrm{d}t}\]
Like before, we can separate that into the expression
\[v\mathrm{d}t=\mathrm{d}x\]
And then into the integral
\[\int_0^t v \mathrm{d}t=\int_{x_0}^x \mathrm{d}x\]

Answer 17

\[v=\frac{ dd }{ dt }\]

Answer 18

So unlike before with a, v is NOT a constant. It's the expression that we found above from our first integral!

\[v=at+v_0\]

So we plug that into our left hand side

\[\int_0^t v \mathrm{d}t=\int_0^t (at+v_0)\mathrm{d}t\]

Answer 19

can you see what that expression gives?

Answer 20

both a and v0 are constants, and the variable you're integrating over is t

Answer 21

\[v=\frac{ 1 }{ 2 }at^2 + v _{0}t\]

Answer 22

your evaluating is perfect! It's not v that's equal to that, however, it's the integral

\[\int_0^t(at+v_0)\mathrm{d}t=\frac{1}{2}at^2+v_0t\]

Which is the left hand side of the original equation.

\[\int_0^t(at+v_0)\mathrm{d}t = \int_{x_0}^x\mathrm{d}x\]
So
\[\frac{1}{2}at^2+v_0t = \int_{x_0}^x\mathrm{d}x\]

What's the right hand side then?

Answer 23

Also, I just noticed that your equation is

\[v_y=.5gt\]

I had misread that and thought it was position, not velocity.

Is there a specific problem that gives that velocity as an answer? (this derivation is still useful to know (and fun to do!), since it's the basis for all kinematics, but I'm unsure where you got your equation from now).

Answer 24

It's for a Physics Lab, where we have to calculate the initial velocity of the y-component of a thrown ball. My physics teacher gave us the formula, but we need to prove it.

Answer 25

We measured the time using a timer.

Answer 26

If you take the derivative of the position function, you would get the velocity function.

Answer 27

That statement is correct. Did you throw the ball straight with the horizontal?

Answer 28

No, it was thrown at an angle, by calculating the x and y initial velocities, i used the formula \[\theta = \left| \tan^{-1} \frac{ v _{y} }{ v _{x}} \right|\] and calculated an angle of about 65 degeres

Answer 29

65º above the horizontal?

Did yo measure either the final velocity of the ball, or the distance it was thrown?

Answer 30

final distance

Answer 31

Here's my draft... If you want to read/skim it... (It still has a lot of typos)

Answer 32

You seem to already have the initial y velocity... (I'm sorry, just trying to figure out what to do).

Answer 33

Yes, I have the formula, but I need to provide a proof for the formula

Answer 34

The proof I tired was wrong, see page 3

Answer 35

That statement as a value for initial velocity doesn't make sense to me - the initial velocity wouldn't be a function of anything. It's just a value.

Answer 36

Your first equation, that resulted in v=at+v0; is what I'm looking for.
The ball reaches it's max at 1/2 t, and it's initial velocity is 0m/s, because it's thrown.

Answer 37

your second statement is not true. If its initial velocity in the y direction was zero, it would only fall and never travel upwards.

Answer 38

It's being accelerated though

Answer 39

Only downwards by the force of gravity.

Answer 40

Hmmm...

Answer 41

And you already used an initial velocity to calculate the angle it was being thrown.

I'm sorry, I just don't know what you're looking for, or what your teacher gave.

Did your teacher really say that the initial velocity in the y direction was
\[v_{yi}=\frac{1}{2}gt\]
?

Or that the velocity in the y direction is equal to that - those are very different statements.

Answer 42

Initial Y velocity, sorry about that

Answer 43

That's what you had said before, and makes very little sense to me at the moment.. hmm...

Initial velocity exists at t=0, so by definition of it being an initial value that expression is zero.

Answer 44

ie,
\[v=v_i+at\]
by definition, an initial velocity is when t=0

\[v=v_i +a(0)\]
\[v=v_i \quad @ t=0\]

Answer 45

The Instructions say, "Calculate initial vertical velocity. This is the acceleration due to gravity, 9.81 meters per second squared or 32 feet per second squared, times half the total time the ball was in the air."

Answer 46

OH! It's not a function of t! t is a delta t. Dealio. k, 1 sec.

Answer 47

Glad that that helped

Answer 48

Definitely was in the wrong headspace.

Okay, so yeah. You basically already did it up top ^_^ Your derivation in the lab didn't work because you used
\[v=d/t\]
which gives a constant velocity over some distance. The ball is being acceleration. So you can either use the integration method from up top, or take the derivative of the position function wrt to t
\[x(t)=v_0t+\frac{1}{2}at^2\]
\[\frac{\mathrm{d}x}{\mathrm{d}t}=v_0 + at\]
\[v(t)=v_0+at\]

Now the fun part!!

So, looking at the graph



What's the y component of the velocity at the top of the parabola?

Answer 49

vy = 0

Answer 50

exactly, so using
\[ t_{Tot} = \text{total time}\]
\[t_{top} = \frac{1}{2} t_{Tot}\]
We have our function
\[v_y(t)=v_{0y}-gt\]
evaluated at time t_top, v_y=0
\[v_y(t_{top})=0\]
\[0=v_{0y}-g(t_{top})\]
\[v_{0y}=g(t_{top})\]
\[v_{0y}=\frac{1}{2}g \ t_{Tot}\]

Answer 51

I'm sorry I made that so complicated before :P Brain death!

Answer 52

No Problem, Thanks alot!!

Answer 53

Very welcome. Good luck with the write up! ^_^

Answer 54

Thanks!