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Football Kinematics

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Homework Statement


Tom throws a football to Wes, who is a distance l away. Tom can control the time of flight t of the ball by choosing
any speed up to vmax and any launch angle between 0◦ and 90◦
. Ignore air resistance and assume Tom and Wes
are at the same height. Which of the following statements is incorrect?
(A) If vmax <√gl, the ball cannot reach Wes at all.
(B) Assuming the ball can reach Wes, as vmax increases with l held fixed, the minimum value of t decreases.
(C) Assuming the ball can reach Wes, as vmax increases with l held fixed, the maximum value of t increases.
(D) Assuming the ball can reach Wes, as l increases with vmax held fixed, the minimum value of t increases.
(E) Assuming the ball can reach Wes, as l increases with vmax held fixed, the maximum value of t increases.

Homework Equations




The Attempt at a Solution


I know that making the launch angle closest to 45 degrees and increasing the horizontal component of the velocity will reduce the time, but I don't know where to go from here.
 

Answers and Replies

  • #2
DEvens
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This is a multiple choice question. And it says "which is incorrect?" So start in checking them.

First, can you check the minimum speed required for the ball to reach? What does that tell you about (A)?

Note that the rest of the statements talk about minimum t and maximum t. Can you work out the min and max times given a vmax and an l? You might have to do some fiddling around there since in principle the max or min could arise from a throwing velocity less than vmax. Once you know how he min and max depend on vmax and l, you should be able to check statements (B) through (E).
 
  • #3
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I found the minimum velocity required to be sqrt(g*L/sin(2*theta)) and assuming theta=45degrees (maximizing sin(2*theta) v=sqrt(gL) making (A) true and not the right answer. The question is how do I work out the min and max times/
 
  • #4
haruspex
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how do I work out the min and max times/
By finding the general expression for the time in terms of l and v.
 
  • #5
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Setting g=10 I got t=L/(v_0*cos(theta) and t=v_0*sin(theta)/5 . I then set them equal to each other and got 5L=v_0^2*sin(theta)*cos(theta). However I don't see how I can derive with a min and max time from that.
 
  • #6
haruspex
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5L=v_0^2*sin(theta)*cos(theta)
You seem to have an extra factor of 2 from somewhere. I think you mean gL=v02*2sin(theta)*cos(theta)= v02*sin(2 theta).
Trouble is, you've eliminated t, which you wanted to find the extrema of. You want t as a function of v, eliminating theta.
 
Last edited:
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The 2 that I have is a exponent of v_0 not a multiplicative factor.
 
  • #8
haruspex
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The 2 that I have is a exponent of v_0 not a multiplicative factor.
OK, I see. That's a hazard of not using superscripts. Conversely, I missed out a factor of 2, now edited in.
Anyway, as I wrote, you want to eliminate theta, not t, to answer the question.
 
  • #9
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How do I eliminate theta? I tried some algebraic manipulation but to no avail.
 
  • #10
haruspex
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How do I eliminate theta? I tried some algebraic manipulation but to no avail.
Cos2+sin2=1.
 
  • #11
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Ok I got v_0^2=5t^2+L^2/t^2. How's that?
 
  • #12
haruspex
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Ok I got v_0^2=5t^2+L^2/t^2. How's that?
Not quite right. Had you kept g instead of substituting 10, you might have noticed the gt2 term has the wrong dimension.
Once you've corrected it, solve for t as a function of v. Yes, I know it's a quartic, but it's actually quite easy to solve.
 
  • #13
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Ok I got gt^4-4v^2+4L^2=0. How is that?
 
  • #14
haruspex
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Ok I got gt^4-4v^2+4L^2=0. How is that?
Same flaw, only worse.
Are you familiar with dimensional analysis? You write M for mass, L for distance, T for time, Q for charge etc. and reduce every term to a combination of these. Ignore constants.
Applying that here, gt^4 has dimension (LT-2)T4 = LT2. v^2 has dimension (LT-1)2 = L2T-2. l^2 has dimension L2. Terms which are to be added together all need to match.
This shows you have mistakes in your algebra.
Your original equations, t=L/(v_0*cos(theta)) and t=2v_0*sin(theta)/g, are dimensionally correct.
 

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