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Projectile motion question

  1. Aug 31, 2013 #1
    1. The problem statement, all variables and given/known data

    The information required is in this picture: http://i.imgur.com/Dt7F9pg.png

    There are four questions based on what is in the picture.

    The questions are:

    1) Calculate how high above the ground the floor of the basket is before the can falls out of the basket.
    2) Determine the vertical component of the can's velocity as it leaves the basket.
    3) Find the time it takes for the can to hit the gouned after it leaves the basket.
    4) Find the time that elapses between the basket touching the ground and the can.

    2. Relevant equations

    Projectile motion equations:
    Vertically:

    [tex]\ddot{y}=a[/tex]

    [tex]\frac{d\dot{y}}{dt}=a\therefore \int d\dot{y}=\int adt\therefore \dot{y}=u\sin\theta +at[/tex]

    [tex]\frac{dy}{dt}=u\sin\theta +at\therefore \int dy=\int \left ( u\sin\theta +at \right )dt\therefore y=u\sin\theta t+\frac{a}{2}t^{2}[/tex]

    Horizontally:

    [tex]\ddot{x}=0[/tex]

    [tex]\frac{d\dot{x}}{dt}=0\therefore \int d\dot{x}=\int 0dt\therefore \dot{x}=u\cos\theta[/tex]

    [tex]\frac{dx}{dt}=u\cos\theta\therefore x=u\cos\theta t[/tex]

    3. The attempt at a solution

    1) ??
    2) 3 m/s
    3) 0.612
    4) The same time?
     
    Last edited: Aug 31, 2013
  2. jcsd
  3. Aug 31, 2013 #2
    1. The balloon has constant speed. So where is it 10s before it hits the ground.
    2. Right
    3. How did you find that if you didn't know how high was it when released? You should better consider it again. (It is also wrong)
    4. Nope... The basket continues falling in the same speed as before. The can is on free fall with initial speed.
    I suggest you make good schematics for every phase of the procedure and the forces involved on the can
     
  4. Aug 31, 2013 #3
    I thought the balloon's velocity would be increasing because of gravity but the question says it is constant. Does that mean the hight would be 30 m?

    For question 3, the answer would become 2.86 s if the height the can is released at is 30 m.

    For number 4, the can would hit the ground first and after 7.14s the balloon would hit it, right?
     
    Last edited: Aug 31, 2013
  5. Aug 31, 2013 #4
    "For question 3, the answer would become 2.86 s if the hight the can is released at is 30 m."
    Check your calculations here. g = 9,81 right? you aree close
     
  6. Aug 31, 2013 #5
    I just recalculated it and got 2.8. Don't know how I got it wrong the first time.
    Thank you!
     
  7. Aug 31, 2013 #6
    You are correct. I forgot about the initial vertical component of the can.
     
    Last edited: Aug 31, 2013
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