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2D Kinematics problem

  • #1
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Homework Statement



A picture of the problem can be seen http://images3a.snapfish.com/232323232fp733;3>nu=52::>379>256>WSNRCG=335;<6;737347nu0mrj".

Original question:

An archer is standing inside a long building whose ceiling is 11 m high. The archer can shoot and arrow at a maximum velocity of 62 m/s, What is the angle that he should shoot at to acheive the greatest range inside the building? How long does the building have to be for the arrow to land inside? How long is it in the air? How fast is it going just before it hits the ground?

Homework Equations



[tex] x_{f}=\frac {a t^{2}}{2} + V_{0}t+x_{o} [/tex]

and other motion equations.

The Attempt at a Solution



Known/Given:

[tex] V_{0}=62 \frac {m}{s} [/tex]

[tex] y_{max}=11m [/tex]

[tex] y_{i}=x_{i}=0 [/tex]

So motion in the x direction is simply going to be:

[tex] x_{f}=V_{0}cos(\theta)t [/tex]

and the y direction should be:

[tex] y_{f} = -\frac {gt^{2}}{2}+V_{o}sin(\theta)t [/tex]

where g is the acceleration due to gravity. Now we know that there is going to be a maximum for y at 11 but we need to get this in terms of 1 variable not two. If I take the derivative of the y equation I get:

[tex] \frac {dy_{f}}{dt} = -gt+V_{0}sin(\theta) [/tex]

we can then solve this to get t in terms of theta, also I believe this is actually a partial derivative:

[tex] \frac {\partial y_{f}}{\partial t} = -gt+V_{0}sin(\theta) [/tex]

[tex] t=\frac {V_{0}sin(\theta)}{g} [/tex]

Now we can substitute this back into the position function to get the position function in terms of only theta:

[tex] y_{f}=v_{0}sin(\theta)(\frac{V_{0}sin(\theta)}{g})- \frac {g}{2}(\frac{V_{0}sin(\theta)}{g})^{2} [/tex]

[tex] y_{f} = \frac {V_{0}^{2}sin^{2}(\theta)}{g} -\frac {g}{2}(\frac {V_{0}^{2}sin^{2}(\theta)}{g^{2}}) [/tex]

[tex] y_{f} = \frac {V_{0}^{2}sin^{2}(\theta)}{g} -\frac {V_{0}^{2}sin^{2}(\theta)}{2g} [/tex]

combine the fractions to get:

[tex] y_{f}= \frac {V_{0}^{2}sin^{2}(\theta)}{2g} [/tex]

now set yf to 11 and solve for theta to get the maximum angle:

[tex] \theta = sin^{-1}(\sqrt{\frac{22g}{V_{0}^{2}}}) \approx 13.6993 [/tex]

Now to figure out how long the building needs to be we should simply need to use the t=F(θ) equation for the new angle and then use that t in the x equation of motion.

[tex] t=\frac {V_{0}sin(\theta)}{g} = \frac {62sin(13.6993)}{9.8} \approx 1.498 seconds[/tex]

[tex] x_{f}=V_{0}cos(\theta)t=62cos(13.6993)(1.498) \approx 90.23 meters [/tex]

I think this is wrong though as the time there is only the time to y max not the total flight time, If you use the time found below for total flight time of the arrow you will get:

[tex] x_{f}=V_{0}cos(\theta)t=62cos(13.6993)(2.996) \approx 180.5 meters [/tex]

To find how long it is in the air we should take the y equation of motion and solve it for t knowing theta and yf (which is 0):

[tex] y_{f}=\frac {1}{2}gt^{2}+V_{0}sin(\theta)t [/tex]

I will use complete the square here because i find it the easiest method for solving quadratic equations:

[tex] 0 =- \frac {1}{2}g(t-\frac{62sin(\theta)}{g})^{2}+ \frac {62^{2}sin^{2}(\theta)}{19.6} [/tex]

solving this for t we get 2.996 seconds.

Finally for how fast is it going when it hits the ground, I think that they want the magnitude of the velocity vector which is:

[tex] \bar{V} = \sqrt{V_{x}^{2}+V_{y}^{2}} [/tex]

and

[tex] V_{x} = V_{0}cos(\theta) \approx 60.236[/tex]

[tex] V_{y} = -gt+V_{0}sin(\theta) \approx -14.68[/tex]

then

[tex] \bar{V}= \sqrt{ 60.236^{2}+(-14.68)^{2}} \approx 62 \frac {m}{s} [/tex]

Is this the correct way to solve the problem?
 
Last edited by a moderator:

Answers and Replies

  • #2
gneill
Mentor
20,781
2,759
I think that I might have gone a slightly different route. Since the objective is to get the maximum range possible, I would go directly to the range equation to see how it behaves when the angle is less than the optimum 45 degrees:

[tex] Range = \frac{v^2}{g} sin(2\theta) [/tex]

and note that for θ < 45° the range increases monotonically with increasing θ. The conclusion is that for the maximum range we want to maximize the launch angle, which means just skimming the ceiling with the arrow.

If the arrow just skims the ceiling, then because horizontal and vertical components of the motion are independent we can fix the y-component of the velocity. The maximum height attained is obtained from conservation of energy:

[tex] \frac{1}{2} m v_y^2 = m \; g \; h [/tex]

[tex] v_y = \sqrt{2 g h} [/tex]

With the total velocity and vy nailed down, then vx and the launch angle follow from Pythagoras and company.
 
  • #3
264
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Fair enough, i am looking into it now. I didn't have my book with me on to read this chapter so I just used what tools I had available at the time.

Is there actually something wrong that I did though? Besides not knowing the range equation?
 
Last edited:
  • #4
gneill
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20,781
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No, you did fine. You even corrected yourself when you started going down questionable paths.

The only minor nit I have, and it's just me perhaps, is that you didn't really justify why the maximum range should be obtained by having the arrow use the maximum available height (11m).
 
  • #5
264
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Well it just made sense to me, haha. If part of Vo is going to be transferred into the y-component (which is obvious) then the max height of the y-component can not be over 11m high, to be technical I think the height is actually the limit of the y component as y approaches 11, not actually equal to but infinitely close.

Also was I correct that the derivative is a partial derivative since theta is held constant?
 
  • #6
gneill
Mentor
20,781
2,759
Well it just made sense to me, haha. If part of Vo is going to be transferred into the y-component (which is obvious) then the max height of the y-component can not be over 11m high, to be technical I think the height is actually the limit of the y component as y approaches 11, not actually equal to but infinitely close.

Also was I correct that the derivative is a partial derivative since theta is held constant?
What would have happened if the ceiling height given was just a bit higher than the maximum height achieved by a 45° launch angle? :devil:

Sure, it's a partial derivative.
 

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