Dart Throwing

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Homework Statement



I have a homwork question that I've been having some trouble with but have come up with a solution.. can someone please verify that i'm on the right path????

The homwork problem is here http://i51.tinypic.com/bbeaf.jpg

Homework Equations




vx= vocos(theta) vy=vosin(theta)

constant vel equations for physics 1

The Attempt at a Solution



Spliting the motion into x-y axis
Horizontal: u= vocosα= 15cosα, a=0, d= 2.5 m
Vertical= u= u= vosinα= 15sinα, a= -9.8, d= 0.05 m
Horizontal: no acceleration so constant velocity time taken to reach to the center
t= 2.5/15cosα = 1/6cosα
Vertical:
-s = +(15sinα)t + 0.5*(-9.8)*t^2
-0.05 = +(15sinα) (1/6cosα) - 0.5*9.8*(1/6cosα)^2
-0.05 = +(2.5tanα) - 4.9/(36sec^2α)
-0.05 = +(2.5tanα) - 4.9(cos^2α) /36
Because angle is very small, cos^2 α ~ 1 .
-0.05= +(2.5tanα) - 4.9/36
α = tan^-1 (0.086)
α = 4.915 degree

now Yoffset
in vertical motion,
Vertical: v=0 ; u= vosinα= 15sinα, a= -9.8, d= ?
02 = (15sin4.92)2 +2*-9.8d
D=8.444 cm (converted from 0.084m to cm)
Yoffset= 13.444 cm
 
Last edited:

Answers and Replies

  • #2
gneill
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It's difficult to follow your math when the divisions have been "lost". It took me a while to figure out that your "16cosα" is really (1/6)*cos(a).

The trig identity is: cos(2θ) = 2cos(θ)2 - 1

So 2cos(θ)2 = cos(2θ) + 1

If 2θ is small, then cos(2θ) ≈ 1, and you find 2cos(θ)2 ≈ 2.

As a result, the angle that I calculate for α is less than half what you found.
 
  • #3
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Sorry for the confusion I fixed some errors in my solution now. I think I have the right answer but am not so sure about the y offset especially..
 
  • #4
gneill
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20,932
2,876
You might want to check your math in these lines:

-0.05= +(2.5tanα) - 4.9/36
α = tan^-1 (0.086)
α = 4.915 degree

Calculating the offset should be a simple matter of trigonometry. If you extend the line through the initial velocity vector to the dart board, and you draw the horizontal line from the launch point to the dart board (note: 5cm above center line), then you have a right angle triangle. You know the horizontal distance, and you've calculated the launch angle in the previous stage, so you can determine the height of the offset above the launch point. Don't forget the 5cm vertical offset of the launch point.
 
  • #5
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First of all thanks for the help!

Okay so for the angle.. i redid my math and have come up with 1.92 degress after I divided by 2.5 and took the tangent inverse in -0.05= +(2.5tanα) - 4.9/36

As for the Y-offset I took the trig approach to the question and came up with 13.38 cm after I solved for the missing side by multiplying tan of 1.92 by 250cm & adding 5cm. Which is close to the value I received for using the constant vel equation initially.

Okay so I think I'm right this time.. did I miss something?

Thanks again!
 
  • #6
gneill
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20,932
2,876
Looks good. The only thing that I might consider is using a more accurate figure for the acceleration due to gravity in the determination of the angle. The question indicated that three figures of accuracy be used, so I'd probably use more for the intermediate calculation values and round the final result.

An acceptable value for g might be 9.807 m/s2
 

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