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Assumption about Projectile Motion in this Question

  • #1
RJLiberator
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Homework Statement


A test rocket is launched, starting on the ground, from rest, by accelerating it along an incline with constant acceleration a. The incline has length L, and it rises at theta degrees above the horizontal. At the instant the rocket leaves the incline, its engines turn off and it is subject only to gravity, g. Air resistance can be ignored. Taking the usual x-y coordinate system, with an origin at the top edge of the incline, a) what is the position vector when the rocket is at its highest point? b) what is the position vector when the rocket is on its way back down and once again at the same height as the top edge of the incline? Symbolic Answer.

Homework Equations




The Attempt at a Solution



My question is this: Is it safe to assume that once you find the time it took for the rocket to reach it's highest point, that you can simply multiply it by 2 to find the time it took to reach y=o again?

I assume this is a yes, but what is throwing my thinking of is that the rocket did have an initial velocity in the y-direction, and when the rocket is at the top of the motion, the rocket no longer has that initial y-velocity coming down, so the time may be skewed.
 

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  • #2
Orodruin
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Homework Statement


A test rocket is launched, starting on the ground, from rest, by accelerating it along an incline with constant acceleration a. The incline has length L, and it rises at theta degrees above the horizontal. At the instant the rocket leaves the incline, its engines turn off and it is subject only to gravity, g. Air resistance can be ignored. Taking the usual x-y coordinate system, with an origin at the top edge of the incline, a) what is the position vector when the rocket is at its highest point? b) what is the position vector when the rocket is on its way back down and once again at the same height as the top edge of the incline? Symbolic Answer.

Homework Equations




The Attempt at a Solution



My question is this: Is it safe to assume that once you find the time it took for the rocket to reach it's highest point, that you can simply multiply it by 2 to find the time it took to reach y=o again?

I assume this is a yes, but what is throwing my thinking of is that the rocket did have an initial velocity in the y-direction, and when the rocket is at the top of the motion, the rocket no longer has that initial y-velocity coming down, so the time may be skewed.
The coming down is the mirror image of the going up (mirrored around the top point). You should compare the initial vertical component with the final one when it hits the ground.
 
  • #3
RJLiberator
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Hm.
So I am wrong to assume that I can simply multiply the time it took to reach height by 2 to get the total time of flight?
Or by mirroring the simulation, am I correct in my assumption?
 
  • #4
PeroK
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Hm.
So I am wrong to assume that I can simply multiply the time it took to reach height by 2 to get the total time of flight?
Or by mirroring the simulation, am I correct in my assumption?
Why don't you try to prove this for all projectile motion under constant gravity?

If a projectile is at height h and moving up at t = 0, reaches its max height at time t_1 and returns to height h moving down at t_2, then is t_2 = 2t_1?
 
  • #5
Orodruin
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Hm.
So I am wrong to assume that I can simply multiply the time it took to reach height by 2 to get the total time of flight?
Or by mirroring the simulation, am I correct in my assumption?
The assumption is correct. If you play a video in the reverse direction at normal speed it will take the same time. If you take projectile motion (neglecting air resistance) and reverse time, it will also look just the same (apart from going in the other direction - but the trajectory is still a parabola).
 
  • #6
RJLiberator
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Thank you for your help - Orodruin.

Perok --> I've tried doing exactly what you stated and heres my results:
T1 = Vi_y/-g
2T1 = 2Vi_y/-g

T2 = (-Vf_y+Vi_y)/g

So therefore, my assumption is correct if i can prove that the final velocity in the y direction is equal to the additive inverse of the initial velocity as that would be 2*T1.
 
  • #7
RJLiberator
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Oh. It is this simple, eh?

Vf_y=Vi_y by the kinematic equation.
It occurs in every situation like this.

That is great.
 
  • #8
PeroK
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Thank you for your help - Orodruin.

Perok --> I've tried doing exactly what you stated and heres my results:
T1 = Vi_y/-g
2T1 = 2Vi_y/-g

T2 = (-Vf_y+Vi_y)/g

So therefore, my assumption is correct if i can prove that the final velocity in the y direction is equal to the additive inverse of the initial velocity as that would be 2*T1.
You might have more luck with:

##s = ut - \frac{1}{2}gt^2##

You can work out when the projectile has 0 vertical displacement. And compare that with the time to reach the highest point, which you calculated above.
 
  • #9
RJLiberator
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A follow up question: When it asks for a 'position vector' what does it want?

I've calculated the max height to be (L*a*sin^2(theta))/g

I am ready to turn it in with that answer, but is that a 'position vector' ?
 
  • #10
SteamKing
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A follow up question: When it asks for a 'position vector' what does it want?

I've calculated the max height to be (L*a*sin^2(theta))/g

I am ready to turn it in with that answer, but is that a 'position vector' ?
A 'position vector' is just what it says: it's the vector drawn from, say, the coordinate origin to the position of the object. There will be two components: either a distance and an angle from the reference point or two components (x and y for 2-D motion).
 
  • #11
RJLiberator
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That is what I was worried about.
But my answer has an angle, and represents a distance. Is this a complete answer?
 
  • #12
SteamKing
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That is what I was worried about.
But my answer has an angle, and represents a distance. Is this a complete answer?
If the angle is that between a line drawn from the top edge of the incline [as stated in the OP, part a)] to the rocket and the x-axis (using the coordinate system specified in the OP), and the distance is measured from the top edge of the incline to the rocket, then, yes, you have your answer for part a). It could also be expressed in terms of the components of the vector, IMO.
 
  • #13
RJLiberator
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Hm. What would I do to express it in terms of the vector component? I am a bit lost there.
 
  • #14
SteamKing
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Hm. What would I do to express it in terms of the vector component? I am a bit lost there.
Your question suggests you are little hazy on the concept of a vector. Am I correct?

In any event, it requires the use of basic trigonometry to find the cartesian components of a vector from its polar form.
 
  • #15
RJLiberator
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I understand the basics of a vector, however, transferring it to polar form will cause me problems. Hm.
 
  • #16
SteamKing
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I understand the basics of a vector, however, transferring it to polar form will cause me problems. Hm.
No, the distance and angle are the polar form. The x and y components are the cartesian form.
 
  • #17
RJLiberator
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Yes, I am very lost on how to convert this symbolic answer to Cartesian coordinates.

Max Height (y) = L*a*sin^2(theta)/g
y = rsin(theta)
y = (La/g)(sin^2(theta))

x = d/2 :/
 
  • #18
SteamKing
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Yes, I am very lost on how to convert this symbolic answer to Cartesian coordinates.

Max Height (y) = L*a*sin^2(theta)/g
y = rsin(theta)
y = (La/g)(sin^2(theta))

x = d/2 :/
You seem to already have cartesian (x,y) coordinates. For x = d/2, it's not clear what d is in terms of the information given in the OP.
 
  • #19
RJLiberator
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d for the total distance, since I solved for max range, i figure I could just divide max range in half which would be equal to the x point at the max height. :p
 
  • #20
SteamKing
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d for the total distance, since I solved for max range, i figure I could just divide max range in half which would be equal to the x point at the max height. :p
Well, that's great, but you still must state what d is in terms of the given quantities.
 
  • #21
RJLiberator
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d = ((4La)cos(theta)sin(theta))/g

d/2 = ((4La)cos(theta)sin(theta))/2g
 
  • #22
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Position Vector for max height: <((4La)cos(theta)sin(theta))/2g, (La/g)(sin^2(theta))>
Position Vector for max range: <((4La)cos(theta)sin(theta))/g, 0>

Is it as simple as this?
 
  • #23
RJLiberator
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Would this be my answer?

Position Vector for max height: <((4La)cos(theta)sin(theta))/2g, (La/g)(sin^2(theta))>
Position Vector for max range: <((4La)cos(theta)sin(theta))/g, 0>
 
  • #24
SteamKing
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Position Vector for max height: <((4La)cos(theta)sin(theta))/2g, (La/g)(sin^2(theta))>
Position Vector for max range: <((4La)cos(theta)sin(theta))/g, 0>

Is it as simple as this?
It's as simple as that.
 
  • #25
RJLiberator
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That does make sense then. I appreciate it.

Thank you.
 

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