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Newton's 3rd Law homework help

  1. Sep 6, 2005 #1
    I'm having trouble with the following problems:
    1) A student throws a water balloon down at his math professor walking on the sidewalk. He drops a balloon from 18.0 meters above the ground when the profesor is 1.00 meter from the point directly beneath the window. IF the professor is 170 cm tall and walks at a rate of .450 m/s does the balloon hit her? If not, how close does it come?
    Given: height= 18 m, v (init)= 0, g= 9.8 m/s2 I dont know where to start after this though for I dont know how to figure out the velocity of the balloon!
    2)A jet aircraft being launched from an aircraft carrier is accelerated from rest along a 94.0 m track for 2.5 sec. a) what is the acceleration of the aircraft, assuming its constant? and b) What is the launch speed of the jet?
    For this one I calculated part a: v= 94 m/2.5 sec = 37.6 m; a= 37.6m/2.5sec = 15 m/s2. However the back of the book states that the answer should be 30 m/s2. And I'm not sure how to go about calculating part b.
    Thanks for any help- i appreciate it!
     
  2. jcsd
  3. Sep 6, 2005 #2
    for (1) Start with calculating the effective striking area from the top view.

    (2) Not enough info to give numbers. What you can do is talk in terms of the relative masses of the carrier and aircraft, and use the law of Action and Reaction (Newton's 3rd Law). Then you can paste together some formulas.
     
  4. Sep 6, 2005 #3

    HallsofIvy

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    What velocity do you want to figure out?? Certainly you can't figure out how far the balloon drops in a certain time by multiplying a velocity by time since the velocity is not constant. Do you know s= -(1/2)gt2+ v0t+ s0? Here v0= 0, s0= 18 m. Set s= 1.7 m, the height of the teacher and solve (-1/2)(9.8)t2+ 18= 1.7. Where will the teacher be at that time?

    I presume you mean that the aircraft travels across the 94 m in 2.5 seconds from an initial speed of 0.
    Your calculation, v= 94 m/2.5 s= 37.s m/s is wrong because that formula holds only for constant speed. That v is the average during the launch, not the final velocity.
    Okay, same formula: s= (1/2)at2+ v0t+ s0. Here a is unknown but t= 2.5, v0= 0, s0= 0 and s= 94. Solve for a. Use that value of a with v= at to find the final launch velocity.
     
    Last edited: Sep 6, 2005
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