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Max height of a projectile equal to range?

  1. Oct 6, 2003 #1
    Hey, I'm having a problem determining the angle necessary for
    the range of a projectile to equal the max height given the velocity
    of the projectile. (the velocity is 53.1m/s)

    I'd imagine it's necessary to set the equation for max height
    equal to the range? Either way, I'm dumbfounded on how exactly
    to obtain the angle of the projectile to have max height = range.

    Any help would be greatly appreciated [I'm just a beginning
    Physics nerd!]
    -Steve
     
    Last edited: Oct 6, 2003
  2. jcsd
  3. Oct 6, 2003 #2
    Is the answer 63.43 degrees?

    I used these equations:

    vox*t = voy*t - .5gt^2
    vf = voy + at
    with a = -g [due to gravity being negative]
    vf = voy - gt
    0 = voy - gt
    since final velocity at peak height is zero
    t = voy/g

    Then subsitute:

    2(vox*voy/g) = 2(voy^2/g) - g*(voy^2/g^2)
    2vox*voy = 2voy^2 - voy^2
    2vox*voy = voy^2
    2vox = voy
    2 = voy/vox

    and since voy = v*sin theta
    vox = v*cos theta

    then 2 = v*sin theta/v*cos theta
    v's cancel

    2 = sin theta/ cos theta
    and since tan = sin/cos
    2 = tan theta

    arc tan 2 = theta
    which is 63.43 degrees

    Is that right?

    Anyone?
     
  4. Oct 7, 2003 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I got a different answer. You may have accidently used the same time for both max height and range!

    vy= v0 sin(θ)- gt= 0 at max height to
    t= v0 sin(θ)/g .

    Also y= v0 sin(θ)t- g/2 t2 is the height so the maximum height is (v02/g) sin2(θ)- (v02/2g)sin2(θ)= (v02/2g)sin2(θ).

    x= v0 cos(θ)t so the range is v0 cos(θ)(2v0 sin(&theta)/g) (Notice the "2". Since we are neglecting air resistance, the motion is symmetrical. The projectile hits the ground in TWICE the time it takes to get to its maximum height.)
    Range= (2 v02/g)sin(&theta)cos(&theta).

    Setting range equal to maximum height,

    (2 v02/g)sin(&theta)cos(&theta)= (v02/2g)sin2(θ).


    v0 cancels (initial speed is not relevant!). We can also cancel the "g" terms. θ= 0 is an obvious solution: if we fire at angle 0 both maximum height and range are 0! Assuming sin(θ) is not 0, we can divide both sides of the equation by that and have

    1/2 sin(θ)= 2 cos(θ) or tan(θ)= 4 so θ= 76 degrees.

    (If you neglect the fact that the projectile hits the ground again in twice the time to max height, you get tan(θ)= 2 and that gives your answer.)
     
  5. Oct 7, 2003 #4
    Ahh, that makes perfect sense now! I thought I had the angle wrong, as it was a rather small angle [and the range and max height were not the same].

    Thanks!
    -Steve
     
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