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Hard Projectile Problem

  1. Jun 27, 2006 #1
    George Washington, first President of the United States of America, is purported to have thrown a silver dollar across the Potomac River from the lawn of his home in Mt. Vernon....The Potomac is approximately 5280ft across at that point. Washington was 6foot2 and threw the coin at an initial velocity of 80 miles/hr.

    What is the farthest distance Washington could have thrown the coin?
    What angle was the coin thrown at to achieve this distance.
    What was the height of the coins' trajectory at its highest point?
    How long was the coin in flight?

    I know how to optimize, but how do I maximize the function accounting for the fact that there are two unknowns. Time and angle.

    Thanks
     
  2. jcsd
  3. Jun 27, 2006 #2

    Doc Al

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    Realize that you have two equations: one for horizontal, one for vertical. Combine them to eliminate time and get horizontal distance as a function of angle only.
     
  4. Jun 27, 2006 #3
    still stuck. How do I combine them. It is not ideal parabolic motion because it doesn't start at the origin.
     
  5. Jun 27, 2006 #4

    arildno

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    It certainly is an exact parabolic trajectory!
     
  6. Jun 27, 2006 #5
    I mean that it is not ideal because it does not start at the origin.
     
  7. Jun 27, 2006 #6

    Kurdt

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    What are your equations of motion and what have you tried to eliminate one of the variables?
     
  8. Jun 27, 2006 #7
    x(t)=117.3cos(a)t
    y(t)=6.167 + 117.3sin(a)t -16t^2
    Distance is in feet. Time is in seconds.

    I've tried to eliminate variables but I have only yielded new variables.
     
  9. Jun 27, 2006 #8

    Kurdt

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    There should be no new variables! Try rewriting t from the x equation and subbing it into the y equation. Its also much easier if you stick to symbols and plug the numbers in at the end.
     
  10. Jun 27, 2006 #9
    "George Washington, first President of the United States of America, is purported to have thrown a silver dollar across the Potomac River from the lawn of his home in Mt. Vernon....The Potomac is approximately 5280ft across at that point. Washington was 6foot2 and threw the coin at an initial velocity of 80 miles/hr."

    Ok, maybe it's just me, but this sounds absurd. I just can't see how a man can throw a silver dollar 1-mile. Nope, don't believe it.
     
  11. Jun 28, 2006 #10

    arildno

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    Besides, there were those pesky Indians about, then, weren't they?
     
  12. Jun 28, 2006 #11

    nrqed

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    From the x equation, [tex] t = {x_f \over v_i cos \theta} [/tex] whre x_f represents the final x position. So
    [tex]
    y_f = y_ i + v_i sin \theta { x_f \over v_i cos \theta} - {1 \over 2} g {x_f^2 \over v_i^2 cos^2 \theta} [/tex]

    This can obviously be simplified (in particular, the relation [itex] sec^2 \theta = 1 + tan^2 \theta [/itex] allows you to rewrite everything in terms of tan theta)
     
  13. Jun 29, 2006 #12

    andrevdh

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    in nrged's equation both [tex]y_f,\ y_i, v_i[/tex] are known quantities. The only unknowns are [tex]x,\ \theta[/tex]. Now we know the x distance travelled is a funtion of the launch angle. We want to know for what launch angle [tex]x[/tex] will be a maximum. Which means that for [tex]\frac{dx}{d\theta}=0[/tex].
     
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