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Help, Kinematics

  1. Jul 24, 2010 #1
    I have trouble solving this question:

    2 projectiles are launched with identical speeds of 30m/s and at angles 40 degrees and 50 degrees with the horizontal, respectively. The difference between the flight times of the 2 projectiles is?

    The answer given to me is 0.754s, round up to 3 significent figures.
     
  2. jcsd
  3. Jul 25, 2010 #2
    Just treat it as two separate projectile problems.

    Velocity is 30m/s so for the first, resolve the velocity horizontally and vertically at an angle of 40 degrees.
    You can then calculate the flight time using just the vertical velocity (v = u + at).

    Do the same for the second and subtract the results.
     
  4. Jul 31, 2010 #3
    [tex]v_{y1}=30sin(40)=19.283, v_{y2}=30sin(50)=22.981[/tex]

    The flying times given that [tex]g=9.81m/s^2[/tex] are [tex]t_{1}=\frac{2v_{y1}}{g}=3.931,t_{2}=\frac{2v_{y2}}{g}=4.685 [/tex]
     
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