Is the Angle of a Projectile at 1/4 of its Flight the Same as When it is Fired?

[jeff12]

My professor did this question in class and I am a little confused. I wrote it down in my notes but I kind of don't understand it.

The question is: Find theta 1/4 of the way through the flight of a projectile in time

He does not give us numbers. Everything has to be solved algebraically.

My question is the angle at 1/4 of the flight the same as the angle of the projectile when it is fired?

And also is the slope of the tangent line at any point on the projectile the same as the slope at that point on the projectile?

 

[PhanthomJay]

Yes, the direction of the projectile at any point in time defines the slope (tangent) of the curve at that point.
Edit: didn't pay attention to your first question. No to that one. The direction of the projectile is changing at each instant.

 

[ehild]

My question is the angle at 1/4 of the flight the same as the angle of the projectile when it is fired?

And also is the slope of the tangent line at any point on the projectile the same as the slope at that point on the projectile?

Welcome to PF!
See the picture: It is the track of a projectile. Is the angle of the tangent of the trajectory is the same everywhere? It would be a straight line then!
Yes, the angle of slope of the tangent line is the same as the angle of slope of the velocity vector of the projectile. The velocity is tangent to the track of motion..
You need the angle the velocity vector encloses with the horizontal direction. How do the horizontal and vertical components of velocity change with time?
What is the time of the whole flight in terms of the initial velocity?

 

[jeff12]

So the slope of the tangent line at that point is equal to the velocity at that point? He doesn't give us numbers. Everything is suppose to be denoted algebraically.

The total flight is t. My notes look something like that. I am confused about the tan.

 

[PhanthomJay]

The slope at time t/4 gives you the direction of the velocity vector at that point. The angle of that slope is calculated as you show : tan theta = Vy/Vx. So you need to find Vy and Vx at that point in time. But that depends on the initial velocity of the projectile at the time it was fired, both it's magnitude and direction, so I assume that was given also as say V_initial at firing angle alpha? Note the firing angle and the angle you are trying to calculate are not the same, so call the firing angle alpha , not theta.

 

[ehild]

So the slope of the tangent line at that point is equal to the velocity at that point? He doesn't give us numbers. Everything is suppose to be denoted algebraically.

The total flight is t. My notes look something like that. I am confused about the tan.

Assume that the initial velocity is of magnitude Vo and it makes the angle α with the horizontal.
You know that the motion of the projectile can be treated as a horizontal and a vertical motion. The horizontal one with constant velocity and the vertical one with constant acceleration. What are the horizontal (x) and vertical (y) components of the initial velocity?
How long is the projectile in air ? Denote it by T.
How do the components of the velocity depend on time?
You are right about the tangent of angle theta. If you have the equations for v_x and v_y, you can write up the equation for tanθ in terms of time, and substitute what you got for t=T/4.

 

[jeff12]

Is tan used to find the angle?

I got to:
tan= sin/cos - gtf/4vcos

 

[ehild]

Is tan used to find the angle?

I got to:
tan= sin/cos - gtf/4vcos

The formula you wrote has no sense.
tan, sin, cos are functions, have no meaning without their argument. The tangent of an angle is its sine over cosine: $$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$$

 

[jeff12]

The formula you wrote has no sense.
tan, sin, cos are functions, have no meaning without their argument.

Yeah, its confusing because he won't give us any numbers. He just did halfway through the problem and said to continue it ourselves.

 

[ehild]

Solve the problem symbolically. Call the initial angle alpha, and denote the angle at t=T_final/4 with theta.

 

[Mister T]

You say he started the problem on the board but didn't finish it? There has to be more information given than what you're telling us. What was the original launch angle? Does the projectile fly across level ground and land at the same height it started from?

I suggest you start over, using the template because it guides you through the start of the process.

Without that template we can't help you.

 

[jeff12]

So since tan=vy/vx

Vx is vcos
Vy is vsin-g(tf/4)

 

[haruspex]

So since tan=vy/vx

Vx is vcos
Vy is vsin-g(tf/4)

You need to distinguish between the angle of the trajectory at launch and the angle at t/4.

 

[jeff12]

Why do I need to do that? I am using tan to find the angle at the 1/4.

 

[haruspex]

Why do I need to do that? I am using tan to find the angle at the 1/4.

You wrote

Vy is vsin-g(tf/4)

What angle is referred to there by "sin"?

 

[jeff12]

You wrote

What angle is referred to there by "sin"?

The angle of 1/4 of projectile.

 

[haruspex]

The angle of 1/4 of projectile.

Then why the -gt/4 term?
(By the way, the projectile is the object. You mean 1/4 of the trajectory.)

 

[jeff12]

Then why the -gt/4 term?
(By the way, the projectile is the object. You mean 1/4 of the trajectory.)

Yes, 1/4 of the trajectory. -gt/4 is 1/4 of the trajectory. He wants it in terms of time.

 

[haruspex]

Yes, 1/4 of the trajectory. -gt/4 is 1/4 of the trajectory. He wants it in terms of time.

What do you mean "-gt/4 is 1/4 of the trajectory"? It is what at 1/4 of the trajectory?

 

[jeff12]

What do you mean "-gt/4 is 1/4 of the trajectory"? It is what at 1/4 of the trajectory?

I actually don't know. He wrote it dovn to find Vy you would use vf=vi+at and then i sub everything into get that. Do you know why he used the formula?

 

[haruspex]

I actually don't know. He wrote it dovn to find Vy you would use vf=vi+at and then i sub everything into get that. Do you know why he used the formula?

Ok, that makes sense. If the initial vertical velocity is v_iy then at time t later the vertical velocity will be v_iy-gt. So if we let T be the total time to land then at 1/4 of the time the vertical velocity, v_y(T/4) will be v_iy-gT/4.

You wrote " Vy is vsin-g(tf/4)". That only makes sense if "vsin" represents the initial vertical velocity, v_iy. But in that case the implied angle in the "sin" must be the launch angle, not the angle at 1/4 of the trajectory time.
So please take much greater care to write complete equations and use different symbols for different variables: discriminate horizontal from vertical, and initial from final from points in between.

 

[Mister T]

I actually don't know. He wrote it dovn to find Vy you would use vf=vi+at and then i sub everything into get that. Do you know why he used the formula?

That's the velocity-time equation for one-dimensional motion. Now that you're looking at two-dimensional motion he's using that formula for each component of the motion.

In particular, for the vertical components of the velocity. He's combining what you learned about one-dimensional motion with what you learned about vectors.

 

[jeff12]

That's the velocity-time equation for one-dimensional motion. Now that you're looking at two-dimensional motion he's using that formula for each component of the motion.

In particular, for the vertical components of the velocity. He's combining what you learned about one-dimensional motion with what you learned about vectors.

If that is the case than why is Vx=Vcos, not like how Vy is found.

 

[haruspex]

If that is the case than why is Vx=Vcos, not like how Vy is found.

Because in the horizontal direction acceleration is zero, so can be omitted.

 

[Mister T]

If that is the case than why is Vx=Vcos, not like how Vy is found.

$v_x=v_{ox}+a_xt=v_o \cos \theta+a_xt=v_o \cos \theta$,
because $a_x=0$.

$v_y=v_{oy}+a_yt=v_o \sin \theta+a_yt=v_o \sin \theta-gt$,
because $a_y=-g$.

What textbook are you using for this course?!

 

[haruspex]

$v_x=v_{ox}+a_xt=v_o \cos \theta+a_xt=v_o \cos \theta$,
because $a_x=0$.

$v_y=v_{oy}+a_yt=v_o \sin \theta+a_yt=v_o \sin \theta-gt$,
because $a_y=-g$.

What textbook are you using for this course?!

To clarify, theta in the above equations is the launch angle. Perhaps we should name this $\theta_0$.

 

[Mister T]

You're correct. The equations should be

$v_x=v_{ox}+a_xt=v_o \cos \theta_o+a_xt=v_o \cos \theta_o$,
because $a_x=0$.

$v_y=v_{oy}+a_yt=v_o \sin \theta_o+a_yt=v_o \sin \theta_o-gt$,
because $a_y=-g$.

 

[jeff12]

You're correct. The equations should be

$v_x=v_{ox}+a_xt=v_o \cos \theta_o+a_xt=v_o \cos \theta_o$,
because $a_x=0$.

$v_y=v_{oy}+a_yt=v_o \sin \theta_o+a_yt=v_o \sin \theta_o-gt$,
because $a_y=-g$.

So I can't simplify it any further right?

The answer would just be tanθ=(vsinθ-g(t/4))/vcosθ
Because I would do inverse tangent to find the angle.

 

[haruspex]

So I can't simplify it any further right?

The answer would just be tanθ=(vsinθ-g(t/4))/vcosθ
Because I would do inverse tangent to find the angle.

Please, please do as I asked and take care to use different symbols for different variables. The equation as written above is obviously wrong. There are two different angles here, the launch angle and the angle at some later time. Likewise, you should use v₀ (or somesuch) for the initial velocity and reserve the unqualified v for the velocity at an arbitrary time t.
Failure to make these distinctions is a major cause of confusion for students, and I strongly suspect is part of your difficulty with this question.

 

[jeff12]

Please, please do as I asked and take care to use different symbols for different variables. The equation as written above is obviously wrong. There are two different angles here, the launch angle and the angle at some later time. Likewise, you should use v₀ (or somesuch) for the initial velocity and reserve the unqualified v for the velocity at an arbitrary time t.
Failure to make these distinctions is a major cause of confusion for students, and I strongly suspect is part of your difficulty with this question.

Okay, so then am I suppose to find the launch angle first and then find the angle at another time after?

 

[haruspex]

Okay, so then am I suppose to find the launch angle first and then find the angle at another time after?

No, you have no way to find out the launch angle. It is one of the inputs to the question. Just leave it as an unknown, $\theta_0$. Rewrite your equation in post #28 using $\theta_0$ and $v_0$ as appropriate... then you are done.

@jeff12, I'm glad you liked my post, but please bring this thread to a satisfactory conclusion by posting that equation using $\theta_0$ and $v_0$ as appropriate.

 

[Mister T]

The best way to think about it is to use these ideas:

1. The vertical component of the velocity $v_y$ will be half of its original value. (Can you explain why?)

2. The horizontal component of the velocity $v_x$ stays the same.

3. $\displaystyle \tan \theta=\frac{v_y}{v_x}$.