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Time of flight of a projectile

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data
    A projectile of mass m is fired from the ground with a speed of 200m/s at an angle to the ground of 30 degrees. Assuming that the air resistance is linear, given by -mλv where v is the projectile's velocity and λ=0.1 s^-1, then find (taking g=10m.s^-2)

    (a)the time of flight (time till it hits the ground again)
    (b)the horizontal distance travelled
    (c)the maximum height reached during the motion

    2. Relevant equations



    3. The attempt at a solution
    Vx = Vo cosα
    Vy = Vo sinα - gt

    X = Xo + Vocosαt
    Y = Yo + V0sinαt - gt^2/2

    Suppose the particle is launched from the origin:

    Xo=Yo=0

    therefore: X = Vo cos αt
    Y = Vo sinαt - gt^2/2

    Eliminating t: t=X/VoCosα

    Substitute into Y: Y=Tanαx - gX^2/2VO^2(Cosα)^2

    Hence: 0=Tan(30X) - 10X^2/2(200)^2(Cos(30))^2

    =Tan(30X) - 10X^2/60000
     
  2. jcsd
  3. Nov 19, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Projectiles

    The angle is "a" so the coefficients are cos a and sin a, "Tanax" means "Tan(a) x", not "Tan(ax)".
     
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