I failure to find an observation related to motion in two dimension projectile motion.
Prove That/Show that
maximum range =4Height

please help me thanks Ahsan Khan

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  1. Professor Hu did it here. Take his derivation for max range, and height, and show that Rmax=4H -

  2. That link did not work
    vertical v = Vo - gt
    at the top, H, v = 0
    t = Vo/g
    then H = Vo t - .5 g t^2
    H = Vo^2/g -.5 Vo^2/g = .5 Vo^2/g
    that t is time to the top, H, which is half the total time in the air
    u = Uo constant horizontal speed
    R = range = Uo *2t = 2 Uo t = 2 UoVo/g
    H/R = .5Vo/2 Uo = (1/4) Vo/Uo
    I suspect that you class has already done the angle for maximum range is 45 degrees above horizontal where Vo = Uo
    H/R = 1/4

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    posted by Damon
  3. Anyway in case you have not done it, for speed S and elevation angle T:
    Uo = s cos T
    Vo = s sin T
    for max height
    t = (s/g) sin T like before
    R = 2 Uo t = 2 s cos T (s/g )sin T
    for max
    dR/dT = 0
    (2 s^2/g)(-sin^2 T + cos^2 T)
    or max range when T = 45 degrees and
    Uo = Vo

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    posted by Damon
  4. i don't know the answer but how 4 the man height varies

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    posted by karan

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