Two Dimensional Motion/Projectile Motion

[Let It Be]

1. I don't have a specific question, however my AP Physics teacher threw these concepts at us, we have a test Wenesday , and I'm getting stressed! I know the general idea of the two, but if someone could give me a more descriptive and concrete explanation I would much appreciate it!


2. ΔX= ViTf + 1/2 ATf^2
H=1/2 gt^2 (free fall)
Vf^2= Vi^2 + 2AΔX
R=2Vi^2 sinθcosθ/g (projectile range) *this is a confusing one to me, and I don't know which questions would need this equation, and how I would use it!


3. I know that for two dimensional, the only difference between this and one dimensional is that now the object isn't going in just one direction, but possibly forwards and up. And for projectile that the horizontal is cos and has no acceleration but constant velocity; and that vertical is sin and has constant acceleration. But if someone could help me more with anything else I should know or any tips it'd be much appreciated!

 

[LawrenceC]

"R=2Vi^2 sinθcosθ/g (projectile range) *this is a confusing one to me, and I don't know which questions would need this equation, and how I would use it!"

Above equation provides the range of a projectile in terms of initial velocity and initial angle from the horizontal from which it was shot. You can derive it by determining the time the projo is in the air in terms of the vertical component of Vi. Then to find the range, R, multiply the horizontal velocity component by the time of flight.

The assumption here is that the horizontal velocity is constant. Your comment in (b) is correct.

 

[Let It Be]

Above equation provides the range of a projectile in terms of initial velocity and initial angle from the horizontal from which it was shot. You can derive it by determining the time the projo is in the air in terms of the vertical component of Vi. Then to find the range, R, multiply the horizontal velocity component by the time of flight.

Yeah, can you dumb this down because I'm still so lost

 

[cupid.callin]

R = \frac{u^2 sin2\theta}{g} ... maximum range

H = \frac{u^2 sin^2 \theta}{2g} ... Maximum height

y = x tan\theta \ - \ \frac{gx^2}{2u^2 cos^2 \theta} ... coordinate's relation at any time ...

u is magnitude of initial velocity
theta is angle with ground at time of launching
x,y ate coordinates at any time ... origin being point of projection

 

[Let It Be]

R = \frac{u^2 sin2\theta}{g} ... maximum range

H = \frac{u^2 sin^2 \theta}{2g} ... Maximum height

y = x tan\theta \ - \ \frac{gx^2}{2u^2 cos^2 \theta} ... coordinate's relation at any time ...

u is magnitude of initial velocity
theta is angle with ground at time of launching
x,y ate coordinates at any time ... origin being point of projection

I'm still pretty lost...how do these equations relate back to the R=2Vi^2sinθcosθ/g?

I appreciate the help so much, but I'm so confused!

 

[cupid.callin]

I'm still pretty lost...how do these equations relate back to the R=2Vi^2sinθcosθ/g?

I appreciate the help so much, but I'm so confused!

use identity: sin2θ = 2 * sinθ * cosθ

 

[Let It Be]

y = x tan\theta \ - \ \frac{gx^2}{2u^2 cos^2 \theta}[/itex] ... coordinate's relation at any time ...

u is magnitude of initial velocity
theta is angle with ground at time of launching
x,y ate coordinates at any time ... origin being point of projection

So then when do I use this equation?

 

[cupid.callin]

So then when do I use this equation?

Well this eqn is kind of mother,

it can solve every problem if used correctly,,, but its quite hectic to solve it as you can see from its size ,,,,

use it if the ques ask something like,,, find height of projectile when its 2 meter from origin horizontally ,,, of course there are other ways to do this ...

 

[Let It Be]

Well this eqn is kind of mother,

it can solve every problem if used correctly,,, but its quite hectic to solve it as you can see from its size ,,,,

use it if the ques ask something like,,, find height of projectile when its 2 meter from origin horizontally ,,, of course there are other ways to do this ...

Yeah, it looks really intense! And sometimes my problems don't have an angle. So you couldn't even use that equation, right?

So then how would you solve the problem you just said (2m from orgin etc...)?

 

[cupid.callin]

just put all values you know and x=2 and get y

or if you know u and theta then you can find time to reach 2m and them vertical distance

 

[LawrenceC]

Yeah, can you dumb this down because I'm still so lost

I'll derive the equation for R, the range of the projectile.

Let's say a projo is shot at an angle of theta with an initial velocity of V. How far will it go neglecting air resistance?

In a problem like this with no air resistance, the horizontal velocity does not change. So if you can determine the time the projo is in the air, you can easily determine its range by multiplying the horizontal component of velocity by the time of flight. So let's write an equation for that.

R = Vh * t
where Vh is the horizontal component of velocity, V*cos(theta). t is time.

Therefore

R = V * cos(theta) * t

We still have an unknown, namely t. To find t, which is the time of flight, we can solve the standard constant acceleration equation for free fall.

0 = Vv * t - 0.5 * g*t^2

where g is acceleration of gravity and Vv is vertical component of velocity and can be written as
Vv = V * sin(theta)

Plugging in and solving for t you get

t = 2 * V * sin(theta)/g

Now plug this function for t into the above equation for range and you get

R = V * cos(theta) * (2 * V * sin(theta)/g)

R = 2 * V^2 * sin(theta) * cos(theta)/g


So if a projo was shot at 100 m/s at an angle of 45 degrees, the range is

R = 2 * 10000 * .707 * .707/9.81 = 1020 m

 

[Let It Be]

I'll derive the equation for R, the range of the projectile.

Let's say a projo is shot at an angle of theta with an initial velocity of V. How far will it go neglecting air resistance?

In a problem like this with no air resistance, the horizontal velocity does not change. So if you can determine the time the projo is in the air, you can easily determine its range by multiplying the horizontal component of velocity by the time of flight. So let's write an equation for that.

R = Vh * t
where Vh is the horizontal component of velocity, V*cos(theta). t is time.

Therefore

R = V * cos(theta) * t

We still have an unknown, namely t. To find t, which is the time of flight, we can solve the standard constant acceleration equation for free fall.

0 = Vv * t - 0.5 * g*t^2

where g is acceleration of gravity and Vv is vertical component of velocity and can be written as
Vv = V * sin(theta)

Plugging in and solving for t you get

t = 2 * V * sin(theta)/g

Now plug this function for t into the above equation for range and you get

R = V * cos(theta) * (2 * V * sin(theta)/g)

R = 2 * V^2 * sin(theta) * cos(theta)/g


So if a projo was shot at 100 m/s at an angle of 45 degrees, the range is

R = 2 * 10000 * .707 * .707/9.81 = 1020 m

Couldn't you also just use
R=2Vi^2sinθcosθ/g
I got the same answer and it seems much easier

 

[Let It Be]

R = \frac{u^2 sin2\theta}{g} ... maximum range

H = \frac{u^2 sin^2 \theta}{2g} ... Maximum height

And then for these two equations, what are they for? Because the first one only has sin to find the maximum range. So would it be maximum range in th Y direction?

And the second...is that for maximum height in the Y direction also?

 

[SammyS]

Height is in the y direction, range in the horizontal direction.

Without using these "magic" equations, do you know the following?

What launch angle gives the maximum height?

What launch angle gives the maximum range (horizontal distance)?​

 

[Let It Be]

Height is in the y direction, range in the horizontal direction.

Without using these "magic" equations, do you know the following?

What launch angle gives the maximum height?

What launch angle gives the maximum range (horizontal distance)?​

No, I don't know that. Just the angle itself? 90° maybe? Could you explain please.

 

[SammyS]

No, I don't know that. Just the angle itself? 90° maybe? Could you explain please.

Certainly, 90° gives maximum height.

How about maximum range ... what angle ?

 

[Let It Be]

Certainly, 90° gives maximum height.

How about maximum range ... what angle ?

Maybe 180°?

Why does that matter? If a problem gave you an angle that was over the limit, what happens?

 

[cupid.callin]

No, I don't know that. Just the angle itself? 90° maybe? Could you explain please.

R = \frac{u^2 sin2\theta}{g} ... maximum range

Hint: For expression of R, what could be value of sin 2θ for R to be maximum

Then what could be θ ...

 

[cupid.callin]

Maybe 180°?

Why does that matter? If a problem gave you an angle that was over the limit, what happens?

180° ?
it will slide on ground ... friction will act and at will slow down pretty soon ...

 

[LawrenceC]

Couldn't you also just use
R=2Vi^2sinθcosθ/g
I got the same answer and it seems much easier

The purpose of the post was to show you where the formula came from as an educational point. It's the same formula you used (R=2Vi^2sinθcosθ/g). Knowing where a formula came from is better than merely plugging in numbers.

 

[JHamm]

You'll probably be served quite well by learning to handle problems without just using a formula somebody gave you, there are only so many formulas you can memorize and only so many problems you can solve using them.