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Are these correct? need a little help

  1. Feb 23, 2010 #1
    Are these right?

    Hi, I just did 2/3 of my assignment for physics and had trouble with the first question. I was wondering if I was correct. Please look over my work.

    1. The problem statement, all variables and given/known data
    3. Two planes fly from Toronto to Philadelphia. Plane A flies via Pittsburgh where as passengers on plane B have a direct flight. Pittsburgh is 350 hm due south of Toronto and 390 km due west of Philadelphia. The airspeed of both planes is 400km/h and a steady wind is blowing from the east at 60km/h.
    a)What direction must the pilot point the plane flying from Toronto to Pittsburgh. Include a vector diagram of velocities.
    b) How long will the entire flight take for plane A assuming a 0.50h layover in Pittsburgh?

    4. A shell is fired from a cliff that is 36m above the horizontal plane. The muzzle speed of the shell is 80m/s and it is fired at an elevation of 25o above the horizontal.
    a) Determine the horizontal range of the shell.
    b) Determine the velocity of the shell as it strikes the ground.

    2. Relevant equations

    v = d/t
    y = ((vi+vf)/2)t
    vf2 = vi2 + 2ay

    3. The attempt at a solution

    3. a) I drew a vector diagram. The hypothesis of the side of the triangle was 400km/h whilst the opposite was 60km/h. The other side (adjacent) turned out to be around 395km/h. This was a right angle triangle.

    I used TANθ=o/a
    TANθ = 60/395
    TANθ = 0.1517
    θ = 9o

    Therefore the plane must travel 400km/h [9o S of W] To reach Pittsburgh. Is this correct?

    b) I used the same diagram and got t = d/v
    t = 350/395
    t = 0.885h

    Vtotal = vnormal-vair
    Vtotal = 400 - 60
    Vtotal = 340km/h

    t= d/v
    t= 390/340
    t= 1.14

    ttotal = 1.14+.5+0.885
    ttotal = 2.5h

    Therefore it would take 2.5h with the 0.5h layover.

    4 a) let up be negative
    vf2 = (-.33.8)2+19.6*36
    vf2 = 1143.08+705.6
    vf2 = 1848.68
    vf2 = 42.99m/s
    vf2 = 43m/s

    t = ((2y)/(vi+vf))
    t = 72 / 76.8057
    t = 0.93s

    d = tv
    d = 80cos25*0.93
    d = 68m

    Therefore the horizontal range of the shell is 68m.

    I used my final velocity for the y-axis to get two sides of the diagram because horizontal does not change.

    c2 = 42.992 = (80cos25)2
    c2 = 1848.68
    c = 42.99m/s

    TANθ = o/a
    TANθ = (80cos25)/42.99
    TANθ =1.6863
    θ = 59.33
    θ = 59

    Therefore the shell hit the ground at 84m/s [59o below the horizontal].

    Are these all correct? Did I do 3 a correct? Where can I improve?
     
    Last edited: Feb 23, 2010
  2. jcsd
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