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Help with Projectile Question

  1. Jun 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A projectile is launched from the top of a 200m tower upwards at a non-vertical angle above level horizontal ground with an initial velocity of 84m/s and rises to a maximum height of 240m above ground level. Determine the angle of projection and time and flights.

    Answer is (19.5 degrees, 4.14 seconds)

    2. Relevant equations

    I know i have to use the V^2 = U^2 + 2as and set V^2 to zero. But because the U is squared and I have 84sin0, im unsure on how to solve for theta.

    3. The attempt at a solution

    I have 2 attempts

    0=84sin0+2*(-9.81)*240
    -2/-9.81x 240 = 84sin0^2
    sqrt (8.49 x 10-4) = 84sin0
    sin-1(0.029/84) = 0
    0 = 0.197

    84sin0^2=sqrt(2-9.81x240)
    sin0^2=68.62/84
    0=54.77
     
  2. jcsd
  3. Jun 2, 2016 #2
    Having trouble understanding your notation (0 instead of [itex] \theta [/itex]) but one mistake I'm seeing right away is the following: in the equation that you're applying, the [itex] s [/itex] stands for displacement. So if the projectile is launched from the top of the 200m tower, and reaches a max. height of 240m; is its displacement really 240m?
     
  4. Jun 2, 2016 #3
    yeah i replaced theta with 0.

    Im not sure, I have used this question to solve for maximum height of the projectile, so I was under the assumption I could use this equation backwards. Im not sure what the actual displacement is. I think that when i do it usually i only take into consideration the height reached above the tower so i shall substitute that in and try again and let you know how i get on. Thank you for your help.
     
  5. Jun 2, 2016 #4
    84 sin^2θ = sqrt (2 x (-9.81) x 40 )
    sin-1(28.01/84)
    = 19.5 degrees

    Thank you
     
  6. Jun 2, 2016 #5
    Nice! Glad you got it :)
    For future reference: the equation refers to the displacement in the motion, since it comes out of the following two equations: [tex] x=x_0+v_0t+\frac{1}{2}at^2 \\ v=v_0+at [/tex]
    Try getting the formula you used from these other two by eliminating the time, it's a worthwhile exercise ;)
     
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