Projectile motion: Bi level equation

  • Thread starter Risker
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  • #1
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This isn't really a homework question, but rather general interest.
I cam across a curious equation the other day,[tex]tan\alpha =\frac{a+b}{a}[/tex]

Where alpha is the launch angle of the projectile b is the height of the platform its being shot onto and a is the distance from the launch to the platform.
For example if you where playing golf and trying to hit the ball over a 1m high fence which was 5m away b=1 and a=5.

I was wondering if anyone knew the proof to this equation, i cannot find it anywhere.
It is not exact but it is a very very close approximation.
Thanks everyone.
 
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Answers and Replies

  • #2
Doc Al
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It is not exact but it is a very very close approximation.
A close approximation to what?

I've never seen this and it doesn't seem to make much sense. What is it's purpose? What if a = b? It gives a zero launch angle.

Where did you see it? Do you have a reference?
 
  • #3
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For example if you were playing baseball, and needed to hit it 21.260 Meters. The launch height is 1.3 meters.
using the formula;
b=-1.3 a=21.260
[tex]\alpha=tan^-1 (\frac{21.260-1.3}{21.260}[/tex]

[tex]\alpha=43.194 (5sf)[/tex]

This is a very close approximation for the launch angle, which is 43 degrees
A colleague showed me this curious equation, she also had no idea why it worked.

Lets try again; You're launching at 43 degrees and needed a range of 21.260, whats required initial height?
[tex]b=(a)tan\alpha -a[/tex]
[tex]b=(21.260)tan(43) -21.260[/tex]
[tex]b=-1.435[/tex] (4sf)
Hence height required is 1.435m which is close to the 1.3 proven by traditional means.

If a=b then there is an obvious limitation.
 
  • #4
Doc Al
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This is a very close approximation for the launch angle, which is 43 degrees
How did you calculate the launch angle of 43 degrees?
 
  • #5
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I assumed a launch speed of 14ms-1 and used equations of motion.
I shall write it up for you in latex tomorrow morning (10 hours from now, I'm Australian.)
 
  • #6
Doc Al
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I assumed a launch speed of 14ms-1 and used equations of motion.
I shall write it up for you in latex tomorrow morning (10 hours from now, I'm Australian.)
No need. As long as you realize that the launch angle depends on the speed of the projectile.
 
  • #7
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Thats why it is so strange that the equation works!!!!
 
  • #8
Doc Al
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Thats why it is so strange that the equation works!!!!
Well, change your assumed speed and see what happens.
 

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