Hard Projectile Problem

  • #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
 

Answers and Replies

  • #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.
 
  • #3
still stuck. How do I combine them. It is not ideal parabolic motion because it doesn't start at the origin.
 
  • #4
arildno
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It certainly is an exact parabolic trajectory!
 
  • #5
I mean that it is not ideal because it does not start at the origin.
 
  • #6
Kurdt
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What are your equations of motion and what have you tried to eliminate one of the variables?
 
  • #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.
 
  • #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.
 
  • #9
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"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.
 
  • #10
arildno
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Besides, there were those pesky Indians about, then, weren't they?
 
  • #11
nrqed
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hippolyta2078795 said:
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.
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)
 
  • #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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