Find the equation y(x) from y(t) & x(t) then find theta max

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In summary, a certain football placekicker can kick the ball a distance d if she kicks it at a 45° angle. Her velocity and height at different points in her trajectory are determined by the equation V = √(Rg/sin2θ).
  • #1
bornofflame
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Homework Statement


A certain football placekicker can kick the ball a distance d if she kicks it at 45°.

a. Find an equation for the velocity of the kick. (You may use the range equation R = (V02/g)sin2θ.)

b. Now imagine she kicks at a different angle θ. Show that a formula for the height y at each horizontal position x from where she kicks is y = xtanθ - (1/2d)x2(secθ)2.
(Hint: This is a projectile motion problem, find x(t), and y(t) just as we did in finding the range equation and put them together to get y(x).

c. Find the angle that gives the maximum height at a given distance x. (Hint: In calculus how did you maximize a value?)

Homework Equations


R = (V02/g)sin2θ
y = xtanθ - (1/2d)x2(secθ)2

The Attempt at a Solution



a. V = ?
V02 = Rg/sin2θ
V0 = √(Rg/sin2θ)
V = V0 + at = √(Rg/sin2θ) + at

Final answer: V = √(Rg/sin2θ) + at

b. x(t), y(t) = ?
x = x0 + v0t + ½at2
Using a triangle where θ is the angle between d and x
p101-test2-redo-2b.png

we can see the following relationships:
x = dcosθ
y = dsinθ
d = x / cosθ
y = (x / cosθ) sinθ = xtanθ This looks like the first half of the equation that we are trying to get to, but I'm not quite sure how to get the other part.

If I look at this in terms of a Pythagorean triangle:
d2 = x2 + y2
y2 = d2 - x2 = (x/cosθ)2 - x2 = x2/sec2θ - x2 = x2(sec2θ - 1) = x2tan2θ = xtanθ

Which kinda makes me feel better as it results in the same thing and gives me the impression that I'm on the right track at least.c. y = xtanθ - (1/2d)x2(secθ)2
Set the derivative to 0 and solve for θ
y = xtanθ - (1/2d)x2(secθ)2 = 0
y' = xsec2θ - (1/2d)x22sec2θtanθ
= xsec2θ - (1/d)x2sec2θtanθ
At first I was thinking that θ would have to be π/2 due to sec2θ being present in both parts of the equation, but the only way for that to work is by taking the limit as θ → π/2 which isn't, I think, the same. So, simplifying further, I got this:
xsec2θ(1 - (x/2d)θtanθ)
Where I still have the d and x haunting me. That is where I'm not sure which way to go. I'm thinking that I may need to solve part b, before I can actually tackle this section.
 

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  • #2
bornofflame said:

Homework Statement


A certain football placekicker can kick the ball a distance d if she kicks it at 45°.

a. Find an equation for the velocity of the kick. (You may use the range equation R = (V02/g)sin2θ.)
Let's start with this part. Can you draw a diagram showing the trajectory of the football and all of the information given in the problem (the angle θ, the angle of the kick, the distance d, the range ##R_0##, the velocity of the kick, and the initial velocity ##V_0##)?
 
  • #3
tnich said:
Let's start with this part. Can you draw a diagram showing the trajectory of the football and all of the information given in the problem (the angle θ, the angle of the kick, the distance d, the range ##R_0##, the velocity of the kick, and the initial velocity ##V_0##)?
P101_test2redo_q2.jpg

Here's what I came up with.
Although, distance should be the arc, while range is the line along the ground, yes?
And since the ball is beginning at rest V0 should be 0 m/s, I think. But I'm not sure because I'm wondering if V0 actually refers to the time just after the ball is kicked. If so, that would change R0, which I have as 0 because of V0.
 

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  • #4
bornofflame said:
View attachment 223328
Here's what I came up with.
Although, distance should be the arc, while range is the line along the ground, yes?
And since the ball is beginning at rest V0 should be 0 m/s, I think. But I'm not sure because I'm wondering if V0 actually refers to the time just after the ball is kicked. If so, that would change R0, which I have as 0 because of V0.
If think you have the the distance d correct in your diagram. You would normally measure the distance kicked as the distance along the ground.
##V_0## must means the speed at time ##t=0## and it is up to you to choose the instant when t is 0. If you say it is before the kick, then you have to figure out the momentum transferred to the ball by the kicker's foot. The ballistic equations don't help you with that, so it's better to skip it and say that ##V_0## is the speed at the instant after the kick when the only force acting on the ball is gravity.
You should show ##V_0## in your diagram. Drawing the diagram and labeling it with everything you know really helps you to figure out the relationships between the variables. I recommend it as the first step in every problem you solve.
 
  • #5
tnich said:
If think you have the the distance d correct in your diagram. You would normally measure the distance kicked as the distance along the ground.
##V_0## must means the speed at time ##t=0## and it is up to you to choose the instant when t is 0. If you say it is before the kick, then you have to figure out the momentum transferred to the ball by the kicker's foot. The ballistic equations don't help you with that, so it's better to skip it and say that ##V_0## is the speed at the instant after the kick when the only force acting on the ball is gravity.
You should show ##V_0## in your diagram. Drawing the diagram and labeling it with everything you know really helps you to figure out the relationships between the variables. I recommend it as the first step in every problem you solve.

Alright. That makes sense. I thought that d was along the ground at first and that is why I made it equal to R, but then I was thinking that the problem was actually referring to the distance traveled through the air. I will fix my diagram to reflect this. You are right. The diagram should be the first thing done. For whatever reason, it's not quite habit yet. I will have to respond after tutoring, however. Thank you for the direction.
 
  • #6
I hope that this is up to snuff. I put in everything that I have and labeled the items in more detail.

I also realized that I probably lost the couple of points on my answer because I failed to consider the value of the angle as 45° and left the θ as a part of my final answer, when it could have been further simplified to V = √(xfg) + at. Should that be further simplified by replacing a with g, since that is the only acceleration in play, non?
p101_test2redo_2a01.jpg

So, with that in mind, instead of uploading multiple pictures, I added the simplification to whiteboard. So, now I have two possible equations for V. I think that, since we don't know what t is at all, the latter equation is the one that should be used. Correct me if I'm wrong.
 

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  • #7
bornofflame said:
I hope that this is up to snuff. I put in everything that I have and labeled the items in more detail.

I also realized that I probably lost the couple of points on my answer because I failed to consider the value of the angle as 45° and left the θ as a part of my final answer, when it could have been further simplified to V = √(xfg) + at. Should that be further simplified by replacing a with g, since that is the only acceleration in play, non?
View attachment 223378
So, with that in mind, instead of uploading multiple pictures, I added the simplification to whiteboard. So, now I have two possible equations for V. I think that, since we don't know what t is at all, the latter equation is the one that should be used. Correct me if I'm wrong.
The diagram looks good, except that you have not labeled the velocity of the kick. When you do that, I think the correct result will become clear to you. Also, you are given the value d for the distance the ball is kicked, but you have not substituted it in your equations.
 
  • #8
I'm not entirely sure what you mean by labeling the velocity. Are you referring to V at the apex when it is 0? Or the final V where it also is 0?

As for d, I was not sure where it went, but after reading what you said and looking back at the diagram, I think I understand: wherever we would write xf or x, we would write d instead which would end up giving the answer V = √(3gd). We could do the same with R as well, since the range is the distance traveled, which would allow us use d = (V02 / g)sin2θ. Or is that only for R? I ask, because I'm not sure where to keep the d and leave the x, since the destination equation for b has both.

By substituting d for R and solving for g after solving for x and y and then substituting x into y, I am able to reach the final equation. The mathematics makes sense to me, the application, however, is still not entirely clear. Again, referring more specifically to when to use d versus x.
p101_test2redo_q2a02.jpg
 

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  • #9
bornofflame said:
I'm not entirely sure what you mean by labeling the velocity. Are you referring to V at the apex when it is 0? Or the final V where it also is 0?
The question asks you for an equation for the velocity of the kick. When does that velocity occur?

bornofflame said:
As for d, I was not sure where it went, but after reading what you said and looking back at the diagram, I think I understand: wherever we would write xf or x, we would write d instead which would end up giving the answer V = √(3gd). We could do the same with R as well, since the range is the distance traveled, which would allow us use d = (V02 / g)sin2θ. Or is that only for R? I ask, because I'm not sure where to keep the d and leave the x, since the destination equation for b has both.
Yes, R, ##x_f## and d all represent the same distance, as you have shown in your diagram. Your answer for V is not correct, though.
 
  • #10
tnich said:
The question asks you for an equation for the velocity of the kick. When does that velocity occur?

I believe that it is referring to velocity from the time that the ball is kicked until the time that it hits the ground; so anytime in between t = 0 and tfinal. If I'm right then an equation for the velocity at any time in between those two times is what is being asked for.
Taking that into account, I've redone my diagram focusing on velocity. I may have gone a little overboard, but this is what I understood for when.
p101_test2redo_q2a04.jpg


tnich said:
Yes, R, ##x_f## and d all represent the same distance, as you have shown in your diagram. Your answer for V is not correct, though.
Ok. Good to know. Thanks.
 

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  • #11
bornofflame said:
I believe that it is referring to velocity from the time that the ball is kicked until the time that it hits the ground; so anytime in between t = 0 and tfinal. If I'm right then an equation for the velocity at any time in between those two times is what is being asked for.
Given the ambiguous wording of the question, I think that is a reasonable interpretation, and now I understand why you did what you did. However, I suspect that what is really being asked for is the initial velocity ##V_0## at the instant after the kick.
I think that if you check some of the points in your diagram, which appears to be correct, against your equation for V, you will find your equation does not give correct answers.
Here is the main problem with your solution approach. You have attempted to use the energy conservation equation ##V^2=V_0^2+2g(H_0-H)## or equivalently, ##\frac12mV^2+mgH=\frac12mV_0^2+mgH_0##. This equation says that sum of kinetic energy (due to motion) and potential energy (due to height) is constant over the trajectory. Do you see why the way you have used it is incorrect?
 
Last edited:
  • #12
tnich said:
Given the ambiguous wording of the question, I think that is a reasonable interpretation, and now I understand why you did what you did. However, I suspect that what is really being asked for is the initial velocity ##V_0## at the instant after the kick.
I think that if you check some of the points in your diagram, which appears to be correct, against your equation for V, you will find your equation does not give correct answers.
Here is the main problem with your solution approach. You have attempted to use the energy conservation equation ##V^2=V_0^2+2g(H_0-H)## or equivalently, ##\frac12mV^2+mgH=\frac12mV_0^2+mgH_0##. This equation says that sum of kinetic energy (due to motion) and potential energy (due to height) is constant over the trajectory. Do you see why the way you have used it is incorrect?

Unfortunately we skipped over the chapter on energy, though we'll be going back to it after we finish momentum; I don't know much at all about kinetic energy and even less about potential right now. I recognize the formula ½mv2 and that kinetic is energy from motion (I think). Even without that, however, what you're saying as far as the energy remaining constant seems quite unlikely. I imagine that as soon as the ball soars through the air it is losing energy that it gained(?) from the force of the kick. Is that why the way that I used it is incorrect?

I chose that equation because at this point we've been told that the equations dealing with constant acceleration, of which this was one, are pretty much interchangeable and it was easy to take the V02 from the range equation and transplant it into this one.
 
  • #13
bornofflame said:
Unfortunately we skipped over the chapter on energy, though we'll be going back to it after we finish momentum; I don't know much at all about kinetic energy and even less about potential right now. I recognize the formula ½mv2 and that kinetic is energy from motion (I think). Even without that, however, what you're saying as far as the energy remaining constant seems quite unlikely. I imagine that as soon as the ball soars through the air it is losing energy that it gained(?) from the force of the kick.
It does lose kinetic energy ##\frac 12mV^2## as it rises but it gains potential energy ##mgH##. The principle of conservation of energy says that the total energy is constant.

bornofflame said:
Is that why the way that I used it is incorrect?
The reason the way you used it was incorrect was that you used horizontal distance ##X## in the equation rather than height ##Y##.

bornofflame said:
I chose that equation because at this point we've been told that the equations dealing with constant acceleration, of which this was one, are pretty much interchangeable and it was easy to take the V02 from the range equation and transplant it into this one.
That's true, you can get to the same result using different ballistic equations.

When you work part b), you will find that you need to know ##V_0## as a function of d. That's why I say you need to be solving for ##V_0## in part a).
 
  • #14
tnich said:
It does lose kinetic energy ##\frac 12mV^2## as it rises but it gains potential energy ##mgH##. The principle of conservation of energy says that the total energy is constant.

The reason the way you used it was incorrect was that you used horizontal distance ##X## in the equation rather than height ##Y##.
Ah. Ok. That makes sense. Since the velocity in the x-direction is not changing at all, but velocity in the y-direction is, right? Even if I had plugged it into the equation using the proper direction, I wouldn't have gotten any further because I was using d = xf.

tnich said:
That's true, you can get to the same result using different ballistic equations.

When you work part b), you will find that you need to know ##V_0## as a function of d. That's why I say you need to be solving for ##V_0## in part a).

As soon as I solved for V0 in terms of d I got what I thought was the answer, until I realized that I have addition in my answer where the desired equation has subtraction. So now I need to figure out where I went wrong with that.
p101_test2redo_q2_05.jpg
 

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  • #15
bornofflame said:
Ah. Ok. That makes sense. Since the velocity in the x-direction is not changing at all, but velocity in the y-direction is, right? Even if I had plugged it into the equation using the proper direction, I wouldn't have gotten any further because I was using d = xf.
As soon as I solved for V0 in terms of d I got what I thought was the answer, until I realized that I have addition in my answer where the desired equation has subtraction. So now I need to figure out where I went wrong with that.
View attachment 223511
In which direction does the force of gravity act?
 
  • #16
Ah. Understood. I didn't realize that I could take the sign of acceleration due to gravity into consideration already. Usually, I am only concerned with it when plugging in numbers.

So that said, for part c I have this:
p101_test2redo_q2c00.jpg

Where I took the derivative and set it equal to zero and solved.
I'm a little rusty on finding maxima but I believe I've done it correctly.
I did it this way because if I take xsecθ out then I would need secθ to equal 0 which can never really happen.
 

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  • #17
bornofflame said:
Ah. Understood. I didn't realize that I could take the sign of acceleration due to gravity into consideration already. Usually, I am only concerned with it when plugging in numbers.

So that said, for part c I have this:
View attachment 223515
Where I took the derivative and set it equal to zero and solved.
I'm a little rusty on finding maxima but I believe I've done it correctly.
I did it this way because if I take xsecθ out then I would need secθ to equal 0 which can never really happen.
You have done just fine except for the last step. The question asks for the angle that gives the maximum height at distance x, not at distance d.
 
  • #18
tnich said:
You have done just fine except for the last step. The question asks for the angle that gives the maximum height at distance x, not at distance d.
I think I can sum up your difficulties in solving this problem as "you need to understand the problem before you can solve it". Drawing a diagram with all of the relevant details (the direction of the acceleration, for example) would go a long ways towards helping you understand the problem. Spending a little more time analyzing the problem statement would help, too. For example, in multi-part problems, you can expect to use the result of part a) in solving part b). That can help you figure out what is needed in part a). That kind of analysis can help a lot on tests when you need to use your time efficiently.

Once you have done that, then start solving equations knowing that you are solving the right equations.

Also, check your results to make sure they make sense. Pick some values of the independent variable for which you know the answer and substitute them into to your final equation to see if you get the right results.
 
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  • #19
tnich said:
You have done just fine except for the last step. The question asks for the angle that gives the maximum height at distance x, not at distance d.

Ah! Thanks for pointing this out.

So then that would be tan-1(d/x) as x gets bigger which means that d/x approaches 0 as x gets larger and we would have tan-10 which would result in θ = x90°, meaning some multiple of 90. Neither 0° or 180° works, however, since that would result in a kick straight off to the side in either direction resulting in little or no height. A kick straight up would seem to fit with the idea of the most height and since the distance traveled in the x-direction is very small, this seems to fit. Hopefully my logic is sound.

tnich said:
I think I can sum up your difficulties in solving this problem as "you need to understand the problem before you can solve it". Drawing a diagram with all of the relevant details (the direction of the acceleration, for example) would go a long ways towards helping you understand the problem. Spending a little more time analyzing the problem statement would help, too. For example, in multi-part problems, you can expect to use the result of part a) in solving part b). That can help you figure out what is needed in part a). That kind of analysis can help a lot on tests when you need to use your time efficiently.

Once you have done that, then start solving equations knowing that you are solving the right equations.

Also, check your results to make sure they make sense. Pick some values of the independent variable for which you know the answer and substitute them into to your final equation to see if you get the right results.

Agreed. I had trouble with this because I didn't fully grasp what the question is asking, which made me feel dumb and probably didn't help; especially after I did well on the first exam. The diagram helped out immensely and I'm definitely kicking myself for not having done it to begin with, it's an invaluable habit that I don't do consistently. Even with the diagram, however, I was missing some bit of knowledge or insight. Thanks for your help and your patience with me, tnich. I appreciate it.
 
  • #20
bornofflame said:
So then that would be tan-1(d/x) as x gets bigger which means that d/x approaches 0 as x gets larger and we would have tan-10 which would result in θ = x90°
I like that you are testing your solution, but you have one detail wrong. ##tan^{-1}(0)## is not 90°.
Try this case (you already know something about it): What happens when ##x=d##?
 
  • #21
When x = d then we have tan-1(1) which evaluates to 45°. So looking at the inside is the important bit. We have d/x. d represents the entire distance that the ball has traveled and x is the point at which V = 0 m/s. So x must be less than d. But if x approaches 0 then we are at a height of 0 or almost 0 which isn't what we want either. So the value of θ is somewhere between 45° and 90°.

This actually ties into a discussion over the problem that I was able to have at school today, I think:
If we take the path of the ball to be symmetrical where the point at which the Vy = 0 is equal to half of d and plug that into our x, that would give us tan-1(d/(d/2)) = tan-1(d*(2/d)) = tan-1(2) = 63.4°. Is that sound?
 
  • #22
bornofflame said:
When x = d then we have tan-1(1) which evaluates to 45°. So looking at the inside is the important bit. We have d/x. d represents the entire distance that the ball has traveled and x is the point at which V = 0 m/s. So x must be less than d. But if x approaches 0 then we are at a height of 0 or almost 0 which isn't what we want either. So the value of θ is somewhere between 45° and 90°.

This actually ties into a discussion over the problem that I was able to have at school today, I think:
If we take the path of the ball to be symmetrical where the point at which the Vy = 0 is equal to half of d and plug that into our x, that would give us tan-1(d/(d/2)) = tan-1(d*(2/d)) = tan-1(2) = 63.4°. Is that sound?
I think you have misunderstood something. You found an equation for ##y(x,θ)##. Then for a given ##x##, you found
##\begin{matrix}argmax\\θ\end{matrix}y(x,θ)=tan^{-1}(\frac dx)##
What that means is for a given ##x##, ##tan^{-1}(\frac dx)## tells you the launch angle to achieve the highest height at a horizontal distance ##x## along the ground from the launch point. You are talking about it as though it would give the highest point in the trajectory, which is not the case.
So for example at distance ##d## from the launch point, the highest height you can achieve is zero. That is consistent with what you know about the range equation. The longest horizontal distance x you can achieve is ##x=d## and that occurs at ##θ=45°##. If you fire at 44° or 46°, the projectile will fall back to a height ##y=0## at ##x < d##. If it continued to fall after that, at ##x=d##, ##y(x)## would be less than 0.
 

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  • #23
Sorry for long time in between posts. All of my upcoming exams are within days of each other and I've been swamped.
On topic: Apparently we weren't looking for an actual angle, which is why it was so difficult, but, rather, the expression that we had already found. So this can all be considered solved.

Thanks again for all your help!
I've definitely been doing better with drawing the diagram either first, or second (after writing out all the known values) and I plan on keeping it up.
 

1. What is the difference between y(x) and y(t)?

Y(x) represents the value of the dependent variable y at a specific value of the independent variable x. On the other hand, y(t) represents the value of y at a specific time t.

2. How can I find the equation y(x) from y(t) and x(t)?

In order to find the equation y(x), you will need to first manipulate the given equation y(t) to express it in terms of x. This can be done by using substitution or algebraic manipulation.

3. What does theta max represent in this context?

In this context, theta max represents the maximum angle or value of the variable theta. This could refer to the maximum value of an angle in a trigonometric function or the maximum value of a variable in a mathematical equation.

4. Can I use any method to find theta max?

The method used to find theta max will depend on the specific equation and variables involved. Some common methods include using derivatives, solving for the maximum value algebraically, or using graphical methods.

5. Is it possible to find y(x) and theta max simultaneously?

Yes, it is possible to find both y(x) and theta max simultaneously. However, the method used to find them may vary depending on the given variables and equation. It is important to carefully consider the given information and use appropriate mathematical techniques to find the desired values.

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