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Height and Range of a projectile

  1. Jul 4, 2009 #1
    A projectile is fired at a speed v0 from and angle [tex]\theta[/tex] above the horizontal. It has a maximum height H and a range R (on level ground)
    Find:
    The angle [tex]\theta [/tex] above the horizontal in terms of H and R

    The initial speed in terms of H, R and g

    and the time of the projectile in terms of H and g.

    Relevant Equations:
    Hmax= [tex]\frac{\left(v0sin\theta\right)^{2}}{2g}[/tex]
    R = [tex]\frac{v0^{2}sin2\theta}{g}[/tex]

    Attempt at a solution:

    From the maximum height equation: v0sin[tex]\theta[/tex]=[tex]\sqrt{2gh}[/tex]
    and from the Range equation: v0cos[tex]\theta[/tex]= [tex]\frac{gR}{2v0sin\theta}[/tex]

    then we have v0cos[tex]\theta[/tex]= [tex]\frac{gR}{\sqrt{2gH}}[/tex]

    Then tan[tex]\theta[/tex]= [tex]\frac{v0sin\theta}{v0cos\theta}[/tex] = [tex]\frac{2H}{R}[/tex]

    so then [tex]\theta[/tex] = tan[tex]^{-1}[/tex][tex]\frac{2H}{R}[/tex]

    Then for the second question, I have v0 = [tex]\sqrt{\frac{gR}{sin2\theta}}[/tex]
    Then I dont know how to convert it to just be in terms of g, H and R

    For the third question I am getting: t = [tex]\frac{2vosin\theta}{g}[/tex]
     
  2. jcsd
  3. Jul 4, 2009 #2
    1. The problem statement, all variables and given/known data

    A projectile is fired at a speed v0 from and angle [tex]\theta[/tex] above the horizontal. It has a maximum height H and a range R (on level ground)
    Find:
    The angle [tex]\theta [/tex] above the horizontal in terms of H and R

    The initial speed in terms of H, R and g

    and the time of the projectile in terms of H and g.

    2. Relevant equations

    Hmax= [tex]\frac{\left(v0sin\theta\right)^{2}}{2g}[/tex]
    R = [tex]\frac{v0^{2}sin2\theta}{g}[/tex]

    3. The attempt at a solution

    From the maximum height equation: v0sin[tex]\theta[/tex]=[tex]\sqrt{2gh}[/tex]
    and from the Range equation: v0cos[tex]\theta[/tex]= [tex]\frac{gR}{2v0sin\theta}[/tex]

    then we have v0cos[tex]\theta[/tex]= [tex]\frac{gR}{\sqrt{2gH}}[/tex]

    Then tan[tex]\theta[/tex]= [tex]\frac{v0sin\theta}{v0cos\theta}[/tex] = [tex]\frac{2H}{R}[/tex]

    so then [tex]\theta[/tex] = tan[tex]^{-1}[/tex][tex]\frac{2H}{R}[/tex]

    Then for the second question, I have v0 = [tex]\sqrt{\frac{gR}{sin2\theta}}[/tex]
    Then I dont know how to convert it to just be in terms of g, H and R

    For the third question I am getting: t = [tex]\frac{2vosin\theta}{g}[/tex]
     
  4. Jul 4, 2009 #3
    You know from this equation, Hmax= [tex]\frac{\left(v0sin\theta\right)^{2}}{2g}[/tex], that

    v0sin(theta) = sqrt(2Hg)

    So plug sqrt(2Hg) into:

    t = [tex]\frac{2vosin\theta}{g}[/tex]

    to get:

    t = 2sqrt(2Hg)/g = 2sqrt(2H/g)
     
  5. Jul 5, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi clarineterr! Welcome to PF! :smile:

    (have a theta: θ :wink:)
    Learn your trigonometric identities …

    you know tanθ in terms of H R and g, so you simply need to express sin2θ and sinθ in terms of tanθ.

    Hint: use sin = cos tan, and cos = 1/sec, and sec2 = tan2 + 1 :wink:
     
  6. Jul 5, 2009 #5
    I got

    [tex]\sqrt{\frac{gR^{2}}{4H}\left(\frac{4H^{2}}{R^{2}}+1\right)}[/tex]

    ??? I dont know if I simplified this right
     
  7. Jul 6, 2009 #6

    tiny-tim

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    Science Advisor
    Homework Helper

    For √(gR/sin2θ) ?

    Yup, that looks good! :biggrin:

    (and now how about your t = 2v0sinθ/g ? :smile:)
     
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