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Projectile Motion Word Problem

  1. Apr 13, 2009 #1
    1. The problem statement, all variables and given/known data

    From the roof of tower 1, a theif throws a money bag to an accomplice on the roof of tower 2, which is just west of tower 1. The two towers are separated by a mall. The defense attorney contends that in order to reach the roof of tower 2, the defendant would have had to throw the money bag with a maximum velocity of no more than 5m/s. Tower 1 is 250 m high, tower 2 is 100m high and the mall is 20m wide. How will you advise the prosecuting attorney?

    2. Relevant equations

    The big 5 equations
    v= d/t
    Cosine Law
    Sine Law
    Pythorean

    3. The attempt at a solution

    First attempt:
    I used calculations assuming a 45 degree angle to the horizontal, then determined time using a big 5 equation. I then used this time to determine that the money traveled a total distance of 31.5 m.

    Second attempt
    d = 1/2 a t^2
    -150 = .5(-9.81) t^2

    5.53 sec is the time to fall from tower 1 to 2

    d = vt
    20 = v(5.53)
    v = 3.61 sec is horizontal speed needed to get to tower 2 in 5.53 sec

    Am I on the right track for either attempt?
     
    Last edited: Apr 13, 2009
  2. jcsd
  3. Apr 13, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Pretty much. If you have 5.5 sec in the air before it reaches 150 m, then to go 20 m in that time is plenty to spare if they can throw at 5 m/s.

    Try beaten by parents as a child for a defense instead.
     
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