What is the trajectory of a projectile and how is it related to parabolas?

[quicksilver123]

Homework Statement

I won't be able to provide the specific problem (in terms of numbers). I was hoping for a conceptual solution.
A ball is thrown. From its initial location, it lands on a shelf a given distance away. I am given the distance to the shelf, the height of the shelf, and the initial angle of incidence.

I am required to find the final displacement in the x direction, as well as the initial velocity of the projectile.

Homework Equations

I have attempted to use one and two dimensional kinematic equations with no lucky.

Specifically:

a trajectory/path equation
(y_f)=tan(theta)(x_f)-(g/(2v_i^2cos²(theta)))*x_f^2

and the one dimensional equation for a free falling body
vyf^2 =vyi^2-2g(yf-yi)

The Attempt at a Solution

I realize that the projectile hits the point of the height of the shelf twice during its trajectory, once during each half of the parabola. Therefore, the y component velocity that I could get from the one dimensional free falling equation seems less than useful to me, as I can't tell those two points apart.
The trajectory/path equation requires both the final x position and the initial velocity, which I do not have. This equation has the upside of being 2D, however, without having the proper 1D equation to complement, I do not think I can use a parametric strategy to solve.

Assistance before the morning?
Thank you.

 

[zexxa]

I'd like to clear up something before helping out with the rest of the question. You said the distance between the thrower and the shelf is given but you're asked to find the displacement in the x-axis? Wouldn't those two be the same or is it on a different plane?

 

[quicksilver123]

Actually, the projectile flies over the shelf, landing some distance beyond it (but still on it!). The shelf is... erm... infinite in length... haha.

 

[NascentOxygen]

the projectile hits the point of the height of the shelf twice during its trajectory, once during each half of the parabola. Therefore, the y component velocity that I could get from the one dimensional free falling equation seems less than useful to me, as I can't tell those two points apart.

The first time the ball attains the level of the shelf the ball has an upwards velocity, the second time it is at that height the ball has a downwards velocity. The ball's vertical velocity during its flight is influenced in a predictable manner by gravity.

If you haven't already drawn a neat sketch showing all this, then you need to do it right now. On that sketch you mark in (i) the information you are given, and (ii) the equations that govern what you are required to find.

 

[quicksilver123]

Well... The second time the projectile is at that height, its velocity is zero. Because it has landed...
So... Really I think I'm just completely stuck and not sure where to go from here. Is the second equation I listed even applicable?
How do I go about doing this? can someone reccomend the correct equation?

 

[zexxa]

No, the velocity is not zero because it does not come to a halt due to gravitational acceleration rather it does so because it 'cannot move anymore'.
It's the same as plotting a parametric graph and solving for x and when you know the value of y. If you haven't drawn the path motion you should do it so as to visualise the question better.

 

[zexxa]

Well I won't blame you if you said that solving it conceptually is hard because using simple kinematics and algebra manipulation it is a huge mess unless you can substitute in numbers at the very start.

The equations that you are using are relevant but you should also consider using $s=ut+½at^2$ as well.
I'll recommend that you try to write t as a function of h , v_i , g , θ and then work backwards again till you have an equation with only 1 unknown (which should be v_i)
Once you've found v_i the displacement in the x-axis should be trivial.

 

[quicksilver123]

sorry, what is "s" and "u"? "t?" There's no time component in this question

 

[zexxa]

s:= displacement of the object from the origin
u:= initial velocity i.e. v_i
I know you don't have a time input in the question but adding that variable makes the equation easier to resolve when you're solving it algebraically.

 

[quicksilver123]

I don't even understand how I would solve it if time isn't given? that's there for a reason right (in the equation)? it has a unit and everything i can't just remove it from the kinematic equation.

the numbers are arbitrary i suppose, but how can i find the answer here? someone mentioned solving it as a parabola, maybe some guidance or a generalized formula for that? its been a long time since high school math..

 

[NascentOxygen]

The only way to solve this is to draw the diagram first, then write down all the equations of motion that you know. The equations are the key here; you can only solve what the equations will allow with the data you have.

If you haven't memorised these, you need to.

 

[PeroK]

I don't even understand how I would solve it if time isn't given? that's there for a reason right (in the equation)? it has a unit and everything i can't just remove it from the kinematic equation.

the numbers are arbitrary i suppose, but how can i find the answer here? someone mentioned solving it as a parabola, maybe some guidance or a generalized formula for that? its been a long time since high school math..

In general, looking at your posts, you are attempting problems beyond your capability. You need some sort of program of revision for things like the maths and the basics of physical motion before you are ready for these problems. You are missing some prerequisites for your studies.

If it's been a long time since high school, how and why are you trying to learn mechanics?

 

[ehild]

Homework Statement

I won't be able to provide the specific problem (in terms of numbers). I was hoping for a conceptual solution.
A ball is thrown. From its initial location, it lands on a shelf a given distance away. I am given the distance to the shelf, the height of the shelf, and the initial angle of incidence.

I am required to find the final displacement in the x direction, as well as the initial velocity of the projectile.

Homework Equations

I have attempted to use one and two dimensional kinematic equations with no lucky.

Specifically:

a trajectory/path equation
(y_f)=tan(theta)(x_f)-(g/(2v_i^2cos²(theta)))*x_f^2

and the one dimensional equation for a free falling body
vyf^2 =vyi^2-2g(yf-yi)

The Attempt at a Solution

The trajectory/path equation requires both the final x position and the initial velocity, which I do not have. This equation has the upside of being 2D,
.

You can solve for the initial velocity Vi if the horizontal coordinate of the landing point is given, using the equation of the trajectory. The distance of the shelf is irrelevant. So you need to clarify, what does it mean that the ball lands on a shelf a given distance away. If it is the distance between points O and P, you can get X from Pythagoras' law.

 

[BvU]

Beautiful picture, Elizabeth ! And: happy birthday

 

[ehild]

And: happy birthday

Thanks, but it is not today.

 

[BvU]

I'm a slow typist... 10 Jan ?

 

[quicksilver123]

Hey!
That was a good post and the diagram looks similar to what I drew. Yes, I am looking for the total x displacement, from O to P.
However, the landing coordinate is not given...

The question does mention that the projectile makes a near miss with the corner of the shelf... is this relevant or helpful?Really I'm stuck because with the given information I see no useful equation for getting the answer.

 

[ehild]

The question does mention that the projectile makes a near miss with the corner of the shelf... is this relevant or helpful?

Yes, it is very relevant and helpful, as you know both the distance and the height of the corner of the shelf, that is, the coordinates of the point P1.

 

[quicksilver123]

Aha!
Thank you!
To be honest, I didn't know what to make of the near miss hint, as it seemed like white noise due to the fact that it does not hit the corner, and therefore the point. I suppose the question maker meant it to be assumed that the miss was arbitrarily close.

 

[ehild]

Aha!
Thank you!
To be honest, I didn't know what to make of the near miss hint, as it seemed like white noise due to the fact that it does not hit the corner, and therefore the point. I suppose the question maker meant it to be assumed that the miss was arbitrarily close.

Yes, it is meant in a Physics problem.
And it is important that you copy the problem text as it is, word by word.

 

[Aufbauwerk 2045]

Hopefully at some point the physics teachers explain that the path of a projectile, such as a cannon ball fired from the Earth's surface, is an ellipse, not a parabola. The parabolic trajectory is an approximation. This would make a nice discussion including Newton's famous thought experiment in which a very fast cannon ball is fired from a very high mountain and achieves Earth orbit.

 

[BvU]

Hopefully at some point the physics teachers explain that the path of a projectile, such as a cannon ball fired from the Earth's surface, is an ellipse, not a parabola. The parabolic trajectory is an approximation. This would make a nice discussion including Newton's famous thought experiment in which a very fast cannon ball is fired from a very high mountain and achieves Earth orbit.

'Some point' being way further on in the curriculum. Parabolas are extremely useful (quadratic equations are the summum of abstraction students can handle at college level -- and often later too ) and match everyday experience, feasible lab experiments etc.