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Finding A Vector Position Function For Projectile Motionh

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Homework Statement


A cannon shoots a ball at an angle θ above the horizontal ground. Neglecting air resistance use newton's second law to find the ball's position as a function of time. (Use the axes with x measured horizontally and y vertically.) Let r(t) denote the balls distance from the cannon. What is the largest possible value of θ if r(t) is to increase throughout the ball's flight? (Hint: using your solution to part (a) you can write down r^2 as x^2 + y^2, and then find the condition that r^2 is always increasing.)


Homework Equations





The Attempt at a Solution


Suppose that the magnitude of the initial velocity is [itex]v_0[/itex]. Since we are neglecting air-resistance, no force opposes the motion directed parallel to the x-axis.

[itex]v_x = v_0 \cos \theta[/itex] (velocity does not change with time)

Integrating, [itex]x(t) = v_0 (\cos)t \theta + x_0[/itex]

The only force acting on the object is the gravitational force.

[itex]F_g = ma[/itex]

rewriting

[itex]a = F_g/m[/itex]

Integrating twice, [itex]y(t) = - \frac{1}{2}gt^2 + v_{y0}t + y_0[/itex], where g = F_g/m and v_yo = v_0 sinθ.

y_0 and x_0 are the initial positions of the object. Define the origin of the coordinate system to coincide with the initial point. Hence, (x_0,y_0) = (0,0)

Now, the vector position function [itex]\vec{r(t}[/itex] can be written as [itex]\vec{r(t)}= (v_0 cos \theta t) \hat{i} + (gt^2 + v_0 sin (\theta) t[/itex]

The magnitude of this is:

[itex]| \vec{r(t)} | = \sqrt{v^2_0 cos^2 \theta - \frac{1}{4} g^2 t^4 + gt^3 v_0 sin \theta + v_0^2 sin^2 (\theta) t^2}[/itex]

-----------------------------------------------------------

Does this appear correct? Also, when it says that you can write r^2 as x^2 + y^2, are they asking me to convert my position into polar coordinates?
 
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Answers and Replies

  • #2
TSny
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See if you can find the (typographical ?) errors in the following:

What is the largest possible value of if r(t) is to...
Integrating, [itex]x(t) = v_0 \cos \theta + x_0[/itex]
[itex]y(t) = \frac{1}{2}gt^2 + v_{y0}t + y_0[/itex]
[itex]\vec{r(t)}[/itex] can be written as [itex]\vec{r(t)}= (v_0 cos \theta t) \hat{i} + (gt^2 + v_0 sin \theta t) \hat{j}[/itex]
[itex]| \vec{r(t)} | = \sqrt{v^2_0 cos^2 \theta 1/4 g^2 t^4 + gt^3 v_0 sin \theta + v_0^2 sin^2 \theta t^2}[/itex]
-----------------------------------------------------------

They are not asking you to work in polar coordinates. But you will want to follow the hint and work with ##r^2## rather than ##r## in order to avoid dealing with the square root.
 
  • #3
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All right, I believe I amended every error. So, the hint is to take the function |r(t)| and square it?
Can't I determine the largest value for θ simply by looking at the vector position function? If θ = pi/2, then there would be no horizontal velocity, and the distance wouldn't increase with time--it would increase, and then decrease. So, the largest value that θ that can attain is one that satisfies [itex]0 \le \theta < \frac{\pi}{2}[/itex]. Isn't this true?
 
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  • #4
TSny
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Just note ##r(t)^2 = x(t)^2 + y(t)^2##. You still have some errors left. Each line that I quoted had at least one error. There's an important sign error associated with the direction of the acceleration of gravity.
 
  • #5
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All right, I think I got them all. Also, I edited post #3, could you possibly check it?
 
  • #6
TSny
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Can't I determine the largest value for θ simply by looking at the vector position function? If θ = pi/2, then there would be no horizontal velocity, and the distance wouldn't increase with time--it would increase, and then decrease. So, the largest value that θ that can attain is one that satisfies [itex]0 \le \theta < \frac{\pi}{2}[/itex]. Isn't this true?
Even if ##\theta_0## is less than ##\pi/2##, ##r## can still end up decreasing through part of the flight unless ##\theta_0## is [STRIKE]large[/STRIKE] small enough.

{EDIT!! Yuck, what I just said is backwards! ##r## will always increase as long as ##\theta_0## is less than a certain value. You need to find that value.}
 
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  • #7
TSny
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It's best not to go back to previous posts and change things too much because it can cause confusion when others are reading through the sequence of posts. Just make the corrections in new posts. (I'm guilty of the same thing.)


[itex]| \vec{r(t)} | = \sqrt{v^2_0 cos^2 \theta - \frac{1}{4} g^2 t^4 + gt^3 v_0 sin \theta + v_0^2 sin^2 (\theta) t^2}[/itex]
One of the terms here is missing some factors of ##t## and two of the terms have the wrong sign.
 
  • #8
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All right, how about I just write r(t) as:

[itex]r(t) = \sqrt{(v_0 \cos (\theta) t)^2 + (- \frac{g}{2} t^2 + v_0 \sin (\theta) t)^2}[/itex]

And squaring,

[itex]r^2 = (v_0 \cos (\theta) t)^2 + (- \frac{g}{2} t^2 + v_0 \sin (\theta) t)^2[/itex]

I am not sure how helpful this is. What should I do next?
 
  • #9
TSny
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You want to show that ##r## is an increasing function of ##t## for all ##t## if ##\theta_0## is small enough. Note that ##r## increases if and only if ##r^2## increases. You have a functional expression for ##r^2(t)##. What is the calculus way of showing that a function is increasing?
 
  • #10
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Would it be taking the partial derivative of r with respect to theta?
 
  • #11
TSny
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No, you want ##r^2(t)## to be an increasing function of time ##t##.
 
  • #12
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Oh, okay, so I take the partial of r^2 with respect to time.. Why does showing r^2 is an increasing function also indicate that r is an increasing function?
 
  • #13
TSny
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The only way for ##r^2## to increase is for ##r## to increase (note ##r## is always positive in our application). And the only way for ##r## to increase is for ##r^2## to increase. A graph of ##r^2## vs ##r## (restricted to positive ##r##) is a good way to see it. So, showing that ##r^2## is increasing with time is equivalent to showing ##r## is increasing with time.
 
  • #14
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All right, I computed the partial derivative of r^2 with respect to time:

[itex]2r \frac{\partial r}{\partial t} = 2v_0 \cos \theta(v_0t \cos \theta) + 2(-gt + v_0 \sin \theta)(- \frac{q}{2} t^2 + v_0t \sin \theta)[/itex] (chain rule)

[itex]2r \frac{\partial r}{\partial t} = 2v_0^2t \cos^2 \theta + 2v_0t \sin^2 \theta - g^2t^3 - 3gv_0t^2 \sin \theta[/itex] (rearranging and simplification)

[itex]2r \frac{\partial r}{\partial t} = 4v_0t - g^2t^3 - 3gt^2 \sin \theta[/itex] pythagorean identity

So, I set [itex]\frac{\partial r}{\partial t} = 0[/itex], and then solve for theta?
 
  • #15
TSny
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You had shown

[itex]r^2 = (v_0 \cos (\theta) t)^2 + (- \frac{g}{2} t^2 + v_0 \sin (\theta) t)^2[/itex]

I think it is easier to square out the expressions and simplifying before taking the derivatives. But of course it should work your way.

All right, I computed the partial derivative of r^2 with respect to time:

[itex]2r \frac{\partial r}{\partial t} = 2v_0 \cos \theta(v_0t \cos \theta) + 2(-gt + v_0 \sin \theta)(- \frac{q}{2} t^2 + v_0t \sin \theta)[/itex] (chain rule)

[itex]2r \frac{\partial r}{\partial t} = 2v_0^2t \cos^2 \theta + 2v_0t \sin^2 \theta - g^2t^3 - 3gv_0t^2 \sin \theta[/itex] (rearranging and simplification)

[itex]2r \frac{\partial r}{\partial t} = 4v_0t - g^2t^3 - 3gt^2 \sin \theta[/itex] pythagorean identity
Your second line looks good except for one sign error.

In going to the last line, it looks like you made a mistake in getting the factor of 4 and also in dropping some factors of ##v_0##.

So, I set [itex]\frac{\partial r}{\partial t} = 0[/itex], and then solve for theta?
You want to find the condition on ##\theta## such that that there will never be a solution for ##t## of the equation ##dr/dt = 0##
 
  • #16
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I believe I fixed the errors, except the one in line two. I don't see how there is sign error.
 
  • #17
TSny
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I believe I fixed the errors, except the one in line two. I don't see how there is sign error.
Should the ##g^2t^3## term be positive or negative?
 
  • #18
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My apologies for being rather tardy in replying. I reworked the solution, and here is what I got:


[itex]r^2 = (- \frac{1}{2}gt^2 + v_0t \sin \theta)^2 + (v_0t \cos \theta)^2[/itex]

Expanding and using the pythagorean identity:

[itex]r^2 = \frac{1}{4}g^2t^4 - gv_0t^3 \sin \theta + 2v_0^2t^2[/itex]

Taking the partial of r with respect to time:

[itex]2r \frac{\partial r}{\partial t} = g^2t^3 -3gv_0t^2 \sin \theta + 4v_0t[/itex]

I really can't see how any value that theta can assume will cause this to be zero.
 
  • #19
TSny
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[itex]r^2 = \frac{1}{4}g^2t^4 - gv_0t^3 \sin \theta + 2v_0^2t^2[/itex]
Check to see if the factor of 2 in the last term is correct.

Taking the partial of r with respect to time:
I really can't see how any value that theta can assume will cause this to be zero.
After taking the derivative with respect to time and setting it equal to zero, you should be able to simplify it to a quadratic equation. Then apply the quadratic formula for solving a quadratic equation.
 
  • #20
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I get[itex]0 = g^2t^3-3gv_0t^2 \sin \theta 2v_0^2t[/itex]. I don't see how this reduces to a quadratic equation.
 
  • #21
TSny
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Divide through by ##t##.
 

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