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Projection Motion Question

  1. Apr 17, 2007 #1
    1. The problem statement, all variables and given/known data
    A 6kg projectile is launced at an angle of 30 degrees to the horizontal and at initial spped of 40m/s. At the top of its flight, it explodes into 2 parts with masses 2 and 4 kg. The fragments move horizontally just after the explosion and the 2kg piece lands back at the launch site.(grav = 9.8)
    Where does the 4kg piece land?

    2. Relevant equations
    Time of flight = 2vsin(theta)/g
    Range = (v^2sin(2*theta))/g

    3. The attempt at a solution

    Using the range equation I got 141.39...m, hence 70.6959...m for the top of the flight.

    Using the time of flight equation I got 4.081..s, hence 2.0408...s for top of flight.

    The 2kg piece travels bak to launch site so it is going to need the same initial speed hence it is -40m/s. p=mv, so p=-80

    therefore the 4kg mass must have a speed of 20m/s (80/4)

    it must take 2.0408 seconds for it to hit the ground again so..
    distance from top of flight = 40.81m

    now add 70.69595 ===> 4kg piece lands 111.5m from launch site.

    I am really terrible with projectile motion :( this is my best attempt however. Please tell me where I have gone wrong. thanks :)
  2. jcsd
  3. Apr 17, 2007 #2


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    Homework Helper

    at the top of its flight, it exploded... means when it exploded, the y-comp of the velocity is zero, leaving only the x-comp which is [tex]v_0 \cos \theta[/tex]. Now, it has to get back to the launch site at a certain time, that time being the time it takes to fall from the position "at the top of its flight" to ground. once you know this time you can work out what the x-comp of the velocity for the 2Kg part (in reverse direction) must be if you know the x-distance between site and position "at the top of flight" .. all of these are quite simple to work out... once you resolve all variables in x- and y- components. note: we are assuming of course that the explosion create two masses that do *not* accelerate in the x-direction, otherwise more info will be needed. by the conservation of momentum before and after explosion gives you the velocity of 4Kg part once you have worked out the velocity of the 2Kg part as outlined above.
  4. Apr 17, 2007 #3


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    oh forgot to mention, the thing was supposedly still travelling in x-direction when it exploded, meaning that you have to be very careful about your frame of reference
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