Confused about projectile motion

  • Thread starter oneplusone
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  • #1
oneplusone
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I don't get how you would attempt problems with 'tilted' projectile motion.
So basically here's a situation:

You are standing on top of a building, and throw a ball down the hill (hill is linear). Find where it lands

Since you are standing on top a building, you have a Δy component, however wherever the ball lands, there's a Δy component there as well.

One idea i thought for these types of problems, is to rotate the 'x' and y axis, such that the x axis is on the slope of the hill. However, since gravity DOESNT act perpendicular in this case, both the x and y component of the initial velocity of the object will have an acceleration. Would this method work?

ALso, is there any easier methods for these types of problems?
 

Answers and Replies

  • #2
mathman
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Your description is not too clear. What is the relationship of the hill to the building?
 
  • #3
oneplusone
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Does this help?
 

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  • #4
ZapperZ
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Does this help?

Are you purposely trying to make the problem more difficult?

Zz.
 
  • #5
Bandersnatch
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Can you show us how you'd solve the problem if the projectile was shot over a flat surface, from the edge of the building?

We'll work from there.
 
  • #6
HallsofIvy
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Do NOT "rotate" any axis. Represent the slope as a straight line through the origin at, say, the base of the building. Find the equation of the parabolic path then where it crosses that straight line.
 
  • #7
MikeGomez
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Do NOT "rotate" any axis.

Why not? It seems like a reasonable approach to me.

Are there any major disadvantages to using this method over the method of intersecting a line with a parabola? Or is it personal preference?
 
  • #8
mikeph
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You're trying to find the intersection points between a parabola (describing the trajectory) and a straight line (the hill).

It is a lot simpler this way than trying to find an expression for the rotated parabola (which would require a parametric expression, f(x,y) = 0) and a flat line.

You'd essentially be making the hard part much harder, and the easy part slightly easier.
 
  • #9
arildno
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1. It is better to define the ground as an explicit function Y(x)
2. Then, find the kinematic position of the ball in terms of functions x(t) and y(t) as solved from Newton's laws of motion.
3. Remember that when the ball hits the ground at some time T, the following equation must hold:
Y(x(T))=y(T)
4. Solve for T, and you're practically finished.
5. Alternatively, you may solve for "t" in terms of x, and represent the curve the ball makes through the air by the function y(x). Then, the intersection point where the ball hits the ground is given by y(x)=Y(x)

Steps 1-4 are roughly as difficult/easy to perform as the procedure outlined in 5
 
  • #10
MikeGomez
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You're trying to find the intersection points between a parabola (describing the trajectory) and a straight line (the hill).

It is a lot simpler this way than trying to find an expression for the rotated parabola (which would require a parametric expression, f(x,y) = 0) and a flat line.

You'd essentially be making the hard part much harder, and the easy part slightly easier.

True. I guess I needed to give more detail. I wouldn't rotate the entire parabola. That would be silly.

I would just rotate the launch point, rotate the gravitation acceleration vector, and rotate the launch velocity. Now the "new" ground line (previously the incline) is horizontal, and we easily find the landing point without using parabola intersecting a line. Then of course we rotate the point back into the original frame.
 
  • #11
mikeph
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Ok, I see what you mean. Yeah, this would work. Never thought about it this way before.
 
  • #12
namanjain
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you just mark along incline x axis and perpendicular to it y axis

acceleration will be gsinθ and gcosθ so make eqn accordingly
 

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