Trajectory: integrating for position

[Oblio]

As usual I did this question I'm working on backwards, doing the section dealing with air friction first. Now realizing I must do the 'first part' and I think I am over thinking again.

With no air resistance, I need the position of a trajectory from Newton's second law.
I know gravity is the only force I'm considering, so I start with:

-mg = ma right?

My prof said when I integrate I'll get combos of sin and cos... I haven't been able to get that.. am I doing something wrong? What steps should I take to integrate this properly?
Thanks!

 

[learningphysics]

Yes, that looks right. What is the initial velocity though... that's probably where the sin and cos come in...

 

[Astronuc]

If one applies Newton's law to a strictly vertical trajectory, then there is no angle, since the velcocity is parallel with the acceleration (force) of gravity.

If one adds a horizontal component of velocity, that is where the angle (with respect to horizontal or vertical) enters the picture.

Air resistance (friction) acts in both orientations (horizontal and vertical), and in opposition to the velocity of the object moving.

See - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

Look at freefall, then vertical launch, then horizontal launch.

 

[Doc Al]

With no air resistance, I need the position of a trajectory from Newton's second law.
I know gravity is the only force I'm considering, so I start with:

-mg = ma right?

Sure: the acceleration of a projectile is g downwards.

My prof said when I integrate I'll get combos of sin and cos...

Perhaps he's talking about sin and cos of the launch angle.

What's the exact problem?

 

[Oblio]

It's not simple to explain because the question refers to 'the projectile of 2.3'.

2.3 compares the trajectory of a projectile subject to linear drag and one in a vacuum. Both have x and y components (not straight up), having the shape of someone hitting a baseball or firing a rocket at 45 degrees.

Vo I assume would be zero then?

Does that help?

 

[Oblio]

The problem is to get the position (x,y) of a projectile with no air resistance (from N2L I'm told) and then compare it to b.) which is considering linear air resistance.

"Assuming there is no air resistance, write down the position (x,y) as a function of t, and eliminate t to give the trajectory y as a function of x."

 

[learningphysics]

It's not simple to explain because the question refers to 'the projectile of 2.3'.

2.3 compares the trajectory of a projectile subject to linear drag and one in a vacuum. Both have x and y components (not straight up), having the shape of someone hitting a baseball or firing a rocket at 45 degrees.

Vo I assume would be zero then?

Does that help?

But if it is being fired... that implies that Vo isn't zero doesn't it? wouldn't horizontal initial velocity be Vocos45, and vertical initial velocity be Vosin45?

 

[Oblio]

But if it is being fired... that implies that Vo isn't zero doesn't it? wouldn't horizontal initial velocity be Vocos45, and vertical initial velocity be Vosin45?

Your right.
45 isn't specified so I hope that's right... no angle is really given.

 

[learningphysics]

Your right.
45 isn't specified so I hope that's right... no angle is really given.

oh... if it isn't specified... then maybe you're just supposed to go with an arbitrary angle $$\theta$$ ?

 

[Oblio]

But acceleration is what I'm starting with, so vcos45 and vsin45 is what I should receive through integration?

 

[Oblio]

oh... if it isn't specified... then maybe you're just supposed to go with an arbitrary angle $$\theta$$ ?

maybe...

 

[learningphysics]

But acceleration is what I'm starting with, so vcos45 and vsin45 is what I should receive through integration?

Do the integration, if that is what is expected in the problem... in both the horizontal and vertical directions... introduce constants when required... then when solving for initial conditions (solving for the constants), that's where you'd put in the vcos(theta), vsin(theta)... don't assume 45 unless it is given.

ie for the part without air resistance... horizontal force is 0... so $$a_x = 0$$

and $$a_y = -g$$

integrate to get vx, vy... then integrate again to get horizontal position, and vertical position.

 

[Oblio]

and then I would integrate those again for position?

 

[learningphysics]

and then I would integrate those again for position?

yes.

 

[Oblio]

F=ma
-mg=ma
integrating:
-mg=m(vcosO +vsinO)

 

[learningphysics]

F=ma
-mg=ma
integrating:
-mg=m(vcosO +vsinO)

careful... deal with the horizontal and vertical separately...

start like this:

ay = -g

now integrate the left side to get vy, what do you get on the right side?

 

[Oblio]

hmm

-mg = m(vsin45)

 

[learningphysics]

hmm

-mg = m(vsin45)

I don't understand how you're getting that.

you start here:

$$-mg = ma_y$$

divide both sides by m

$$a_y = -g$$

now integrate...

 

[Oblio]

I don't understand how you're getting that.

you start here:

$$-mg = ma_y$$

divide both sides by m

$$a_y = -g$$

now integrate...

$$v_{y}sin45 = -g$$ ?

 

[learningphysics]

$$v_{y}sin45 = -g$$ ?

don't worry about the initial velocity for now... you want to integrate -g, with respect to time...

suppose you had this question instead:

$$\frac{dy}{dx} = -9.8$$

Find y.

 

[Oblio]

y= -9.8t

 

[learningphysics]

y= -9.8t

yeah... but you need to introduce a constant...

when you integrate

$$a_y = -g$$

you get:

$$v_y = -gt + C$$

 

[Oblio]

right, right. sorry, been a while since Calc. classes.

 

[Oblio]

so, to get position, I would end up with

-gt^2 +Ct + C = y ?

 

[learningphysics]

so, to get position, I would end up with

-gt^2 +Ct + C = y ?

almost careful... don't use the same constant

y = (-1/2)gt^2 + Ct + D

now you can solve for C and D... you know that

vy = -gt + C
y = (-1/2)gt^2 + Ct + D

at t=0, vy = vsin(theta)
I assume, that at t = 0, y = 0.

What is C? What is D?

 

[Oblio]

so, by eliminate t they mean set it to 0?

 

[Oblio]

Nm, that's not what you meant.

 

[learningphysics]

so, by eliminate t they mean set it to 0?

No. don't worry about that part yet... first we need the constants C and D...

 

[Oblio]

D = 1/2 gt^2 - C +y

C = 1/2gt +y - D

 

[learningphysics]

D = 1/2 gt^2 - C +y

C = 1/2gt +y - D

Start with this:

vy = -gt + C

now... you know that at t = 0, vy = v0sin(theta)... substitute t and vy into the equation... and solve for C.