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Homework Help: Show that the horizontal range is 4h/tan(theta)

  1. Jan 27, 2014 #1
    1. The problem statement, all variables and given/known data

    A projectile launched at angle Θ to the horizontal reaches maximum height h. Show that its horizontal range is 4h/ tan Θ.

    3. The attempt at a solution

    tapex = [itex]vi sin Θ/g[/itex]
    tfull trajectory = [itex]2vi sin Θ/g[/itex]
    h(tapex) = h = [itex]vi^2 sin^2 Θ/2g[/itex]

    x(tfull trajectory) = x = [itex]\frac{vi^2 sin (2Θ)}{g}[/itex]
  2. jcsd
  3. Jan 27, 2014 #2
    Consider the ratio of range to max height.
  4. Jan 27, 2014 #3

    I did. The expression for the full range and heigh, h is in the op.
    But I'm unable to reduce them to the appropriate wxpression
  5. Jan 27, 2014 #4
    Show what you get and how you get that.
  6. Jan 27, 2014 #5
    I worked out hapex = vi ^2 sin^Θ and xfull range = vi^2(2Θ)/g
    The above was obtained by substituting t/2 intp the y-displacement and t into the x-displacement.
  7. Jan 27, 2014 #6
    As I said. Consider the ratio of range to max height.
  8. Jan 27, 2014 #7
    I did.

    The horizontal range is 2vi^2 sinΘcosΘ.
    I simplified 4h/tanΘ to 2vi^2sinΘcosΘ/g.
    I suppose this is a sufficient condition for the proof?
  9. Jan 28, 2014 #8
    I cannot see your analysis of the ratio of range to max height. Which is strange, because you have found a formula for range, and a formula for max height. All you need is to divide one by another.
  10. Jan 28, 2014 #9

    I failed to obtain the required equation. Why do we need the ratio? The question has not asked for that.

    My method:
    I found x = vi^2 sin(2Θ)/g = 2vi^2 sinΘcosΘ
    I reduced 4h/tanΘ to = 2vi^2 sinΘcosΘ, where h = vi^2 sin^2Θ/2g and tan Θ = sinΘ/cosΘ.
  11. Jan 28, 2014 #10
    You are required to prove that $$ x = {4h \over \tan \theta} $$ That means $$ {x \over h} = {4 \over \tan \theta} $$

    Regarding your "failure", how are ## \cot ## and ## \tan ## related?
  12. Jan 28, 2014 #11

    cot Θ= 1/tanΘ
    It never occurred to me I had to perform a x/h ratio. But anyway, answer found.
    Last edited: Jan 28, 2014
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