Projectile motion help needed !

  • Thread starter elphin
  • Start date
  • #1
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projectile motion help needed!!!!!!

Homework Statement



If v1, v2, v3 are the velocities at three points A, B , C of the path of a projectile, where the inclinations to the horizon are α, α-β, α-2β and if t1, t2 are the times of describing the arcs AB and BC respectively, prove that

v3*t1 = v1*t2 and 1/ v1 + 1/v3 = (2*cos β)/v2


Homework Equations





The Attempt at a Solution



v1cos α = v2cos (α-β)= v3cos (α-2β) - (horizontal component same always)

and

v2sin(α-β) = v1sin α – g*t1 (&) v3sin (α-2β) = v2sin(α-β) – g*t2

and now I am stuck .. are there any more equations that I am missing or is it a bad case of trigonometric manipulation.. I tried trigonometric manipulation .. but ended up getting weirder equations
 

Answers and Replies

  • #2
302
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I think it's a case of choosing the right trigonometric manipulations. I got the second relation-

Using your formulas,

[itex]\frac{1}{v_{1}}+\frac{1}{v_{2}}=\frac{cos(\alpha)}{v_{2} cos(\alpha-2\beta)}+\frac{cos(\alpha-\beta)}{v_{2} cos(\alpha-\beta)}[/itex]

[itex]cos(\alpha) + cos(\alpha-2\beta) = cos(\alpha) + cos(\alpha)cos(2\beta) - sin(\alpha)sin(2\beta)[/itex]
[itex]cos(\alpha) + cos(\alpha-2\beta) = cos(\alpha)(1+cos(2\beta)) - 2sin(\alpha)sin(\beta)cos(\beta)[/itex]
[itex]cos(\alpha) + cos(\alpha-2\beta) = cos(\alpha)(1+2cos^{2}(\beta)-1) - 2sin(\alpha)sin(\beta)cos(\beta)[/itex]
[itex]cos(\alpha) + cos(\alpha-2\beta) = 2cos(\beta)(cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)) = 2cos(\beta)cos(\alpha-\beta)[/itex]

Resubstitute and get the required result. However I don't see any physical significance to this..
 

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