# Homework Help: Tangent Line on Projectile

1. Oct 8, 2015

### jeff12

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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?

Last edited: Oct 8, 2015
2. Oct 8, 2015

### 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.

Last edited: Oct 8, 2015
3. Oct 8, 2015

### ehild

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?

4. Oct 9, 2015

### 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.

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Last edited: Oct 9, 2015
5. Oct 9, 2015

### 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.

6. Oct 10, 2015

### ehild

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 vx and vy, you can write up the equation for tanθ in terms of time, and substitute what you got for t=T/4.

7. Oct 11, 2015

### jeff12

Is tan used to find the angle?

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

8. Oct 11, 2015

### ehild

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: $$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$$

9. Oct 11, 2015

### jeff12

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

10. Oct 11, 2015

### ehild

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

11. Oct 11, 2015

### 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.

12. Oct 12, 2015

### jeff12

So since tan=vy/vx

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

13. Oct 12, 2015

### haruspex

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

14. Oct 12, 2015

### jeff12

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

15. Oct 12, 2015

### haruspex

You wrote
What angle is referred to there by "sin"?

16. Oct 12, 2015

### jeff12

The angle of 1/4 of projectile.

17. Oct 12, 2015

### haruspex

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

18. Oct 12, 2015

### jeff12

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

19. Oct 12, 2015

### haruspex

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

20. Oct 12, 2015

### jeff12

I actually dont know. He wrote it dovn to find Vy you would use vf=vi+at and then i sub everything in to get that. Do you know why he used the formula?

21. Oct 12, 2015

### haruspex

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

You wrote " Vy is vsin-g(tf/4)". That only makes sense if "vsin" represents the initial vertical velocity, viy. 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.

22. Oct 12, 2015

### Mister T

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.

23. Oct 12, 2015

### jeff12

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

24. Oct 12, 2015

### haruspex

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

25. Oct 12, 2015

### Mister T

$v_x=v_{ox}+a_xt=v_o \cos \theta+a_xt=v_o \cos \theta$,
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?!