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Projectiles problem, help needed

  • #1
Here is the problem (Q3 (A));
ImageUploadedByPhysics Forums1423676612.394766.jpg


Here's what I have so far;
ImageUploadedByPhysics Forums1423676690.906861.jpg


Any help would be appreciated, thank you!
 

Answers and Replies

  • #2
RUber
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It appears you made an algebra error when you went from ## 21 \cos \alpha = \frac {30}{ t} ## to ## \cos \alpha = \frac {630}{ t} ##.
Also, I would first find your time of impact based on your horizontal component.
With that time of impact, you can make an expression that is in terms of alpha only.
 
  • #3
It appears you made an algebra error when you went from ## 21 \cos \alpha = \frac {30}{ t} ## to ## \cos \alpha = \frac {630}{ t} ##.
Also, I would first find your time of impact based on your horizontal component.
With that time of impact, you can make an expression that is in terms of alpha only.
Yes I realised this after I posted, I'll try that, thanks!
 
  • #4
RUber
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I made a substition of x = cos^2(alpha) and manipulated it into a quadratic equation to get the two solutions.
part b comes right out of part a if you have already solved for time of impact.
 
  • #5
TSny
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Aaron,

Something else to note: If you have the expression ##\tan \alpha = \frac{A}{B}##, then you cannot conclude that ##\sin \alpha = A## and ##\cos \alpha = B##.
 
  • #6
I made a substition of x = cos^2(alpha) and manipulated it into a quadratic equation to get the two solutions.
part b comes right out of part a if you have already solved for time of impact.
Could you elaborate on this a bit? Not quite sure what you mean..
 
  • #7
RUber
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Do you have an expression that can be written as some trig functions of alpha = 0?
Use sin = sqrt(1-cos^2) to convert everything in terms of cos.
 

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