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Homework Help: Hitting a moving target below when launched 100 m above at a 45 degree angle

  1. Dec 23, 2012 #1
    i'm having difficulty trying to figure out a problem i thought of.

    a projectile is launched upwards at a 45 degree angle 100 m above a surface. below is a target whose horizontal distance is 5 m from the launch point. just as the projectile is launched, the target moves horizontally at a constant 9 m/s.

    what would the velocity of the launch have to be in order to hit the target below?

    so far i have come up with:

    5 + 9t = vcos45t

    for t = time i have:

    (see attachment)

    but i'm having difficulty knowing how to incorporate it properly into the equations.

    i've reduced the time to:

    √([.02548v2 + 100]/4.9) + (.07215v)

    is there an easier way to go about solving this?

    any help would be greatly appreciated.

    thank you

    Attached Files:

    • 01.bmp
      File size:
      40 KB
  2. jcsd
  3. Dec 23, 2012 #2


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    Staff: Mentor

    I don't think there is an easier way.
    I would try to use g instead of its numerical value and the analytic value for sin(45°) as long as possible, that avoids decimal numbers.
  4. Dec 23, 2012 #3

    i came up with a velocity of 13.985 m/s at a time of 5.638 seconds

    thanks for your help...now i can sleep soundly :-)
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