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Homework Help: Projectile Angle and Distance problem

  1. May 30, 2016 #1
    1. The problem statement, all variables and given/known data

    A projectile was launched from the top of a 100m tower upward at an angle theta to the horizontal with an initial velocity of 36m/s. The time of flight of the projectile was 8 seconds.

    i) The angle of projection
    ii) The maximum height reached by the projectile

    2. Relevant equations


    3. The attempt at a solution

    I am used to solving these questions when i am not given time but given both the horizontal and vertical distances. At which I use the quadratic formula to solve for theta. However in this case I can not do that as when i split my horizontal and vertical components and apply the above equations I am left with these.



    I have tried rearranging the vertical formulas to solve for theta

    301.42 = 36*sin(?)
    sin-1(36/301.42) = 6.85 degrees

    I really don't think that the answer is correct. The lecture who set the question will not give any answers.
  2. jcsd
  3. May 30, 2016 #2


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    You have only the 100 being divided by 8. But is that correct?
  4. May 30, 2016 #3
    Yeah. Only the 100 divided by 8. Im thinking that maybe 6.85 degrees is the correct angle for a projectile to fall from 100 metres in 8 seconds.
  5. May 30, 2016 #4


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    Check the algebra on this. I don't think it's correct to have just the 100 divided by 8.
  6. May 30, 2016 #5
    Im not exactly sure on the rerrangment to solve for U. As i thought my 1st rearrange was correct.

    I tried the rearrange dividing the whole thing by 8 = sin-1(26.74/36) = 47.96 degrees.

    I tried the rearrange dividing by 16 from 1/2*8 under everything = sin-1(36/60.76) = 36.33 degrees
  7. May 30, 2016 #6
    s = ut + 1/2at^2
    ut = s - 1/2at^2
    u = (s - 1/2at^2)/t
    36sin(?) = (-100-1/2*-9.81*64)/8

    Sin(?)= 26.74/36
    Sin-1(26.74/36) = (?)
  8. May 30, 2016 #7


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    That looks correct.
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