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Physics Honors-Falling Bodies

  1. Nov 27, 2013 #1
    1. The problem statement, all variables and given/known data

    A boy launches a toy helicopter from the ground beside a building. It accelerates upward at 15 m/s^2. At the same time, another boy throws a rock down from the top of the building, which is 60 m above the ground. the helicopter and the rock collide at a height of 30 m. How fast was the rock thrown?

    Going of the problem:
    a=15 m/s^2 a=-8.8 m/s^2
    x=30 m x=30 m
    xinitial= 60m xinital=60 m
    v? v=?



    2. Relevant equations

    v^2=2aΔx
    3. The attempt at a solution

    I tried averaging the numbers and used the above equation to get the average velocity. I got ≈12.5 m/s. However, I'm not sure if average velocity is exact enough for this problem. Does anyone have an alternate suggestion?

    Thanks!!
     
  2. jcsd
  3. Nov 27, 2013 #2
    What numbers did you average? And why do you have to do it? There is no question about an average speed.
    What equations of motion have you used?
     
  4. Dec 2, 2013 #3
    Hi Nasu,

    Sorry this reply took so long! I believe I averaged the two accelerations and the two velocities to plug into this equation: v^2=2aΔx

    I was trying to find the velocity of the rock, but I'm not sure how to go about it. I thought I'd use the above equation just because it was one I'm fairly familiar with, and I hoped it would somehow prove useful.

    Thanks!
     
  5. Dec 3, 2013 #4
    To expand on nasu's hint..

    What other equations do you know that might be relevant? While it is great if you instantly know the correct equation to use, when unsure a list of equations that might be useful along with a table of knowns and unknowns for each equation can often illuminate how to solve problems. Questions are often structured so that at least one of the equations that you might use has just a single unknown. Once solved that result feeds into other equations to produce further results until the motion is completely described and the problem completed.
     
  6. Dec 3, 2013 #5
    Even before equation, it is essential to understand what is going on. A drawing may be a great help.
    When you have problems with two moving bodies meeting or colliding, this means that they will be in the same position at the same time.
    So you should think about equations involving these two quantities. And using the condition for collision: same time and same position.
    These conditions will give the unknown quantity (initial velocity).

    The equation you wrote does not apply for the body thrown from the top. The equation is actually
    [itex]v_f^2=v_i^2+2a\Delta x[/itex]
    It may reduce to the one you wrote if either initial or final speed is zero. Which is not the case here.
     
  7. Dec 3, 2013 #6
    Thanks to both of you for explaining! I looked on an equation sheet but I couldn't tell which would be most helpful. There is a drawing on my paper, thankfully. Drawing is hard. xD nasu, thank you for the equation! For some reason selecting the equation seems to be the most challenging part of solving problems. I am familiar with the one you gave, but I'm not sure how to use it with two different numbers for each value. As far as I know, we have two accelerations:
    15 m/s^2 and -9.8 m/s^2

    two final distances:
    30 m and 30 m

    And two initial distances:
    60 m and 60 m

    The acceleration numbers are different, but everything else is the same. If I am right about there being two completely different accelerations, how do I go about using this formula?

    Thanks again!!!
     
  8. Dec 3, 2013 #7
    I did not say to use that formula. I just showed the complete form to explain why the one you wrote originally does not even apply here.

    You need an equation that relates position and time. And write that equation for each one of the two moving bodies.
     
  9. Dec 4, 2013 #8
    nasu,

    Oh, I must have misunderstood the wording. Thanks! :-)

    Would x=x0+v0t+1/2at^2 be the easiest formula to use?
     
  10. Dec 4, 2013 #9
    Yes, this is the general formula.
    Now you write for the actual bodies in the problem.
    You have to specify x0, v0, a for each of the two objects.
    And you will have two equations with two unknowns.
     
  11. Dec 5, 2013 #10
    x=x0+v0t+1/2at^2 x=x0+v0t+1/2at^2

    Putting in the numbers: Putting in the numbers:
    30m=60m+v0t+1/2(15 m/s^2)t^2 30=60m+v0+1/2(-9.8m/s^2)t^2

    Simplifying as much as possible: Simplifying as much as possible:
    30m=60m+v0t+(7.5m/s^2)t^2 30m=60m+v0t+(4.9m/s^2)t^2

    I'm not really sure what to do next, because I'm only supposed to have an unknown for each equation, correct? Does v0=0 for both because it says the boy "launched" the helicopter and another boy "threw" the rock, implying that they just started?
     
  12. Dec 5, 2013 #11
    Ok, this is for the body coming down, right?
    Taking x and xo positive means that we measure the distance from the ground up. So up is positive.
    Then vo and a are negative, they are pointing down. And a is not 15m/s^2. This value is for the helicopter going up.

    Make these corrections and you will have an equation with two unknowns: t and vo.

    Then write the similar equation for the helicopter. And you will have a second equation.
     
  13. Dec 6, 2013 #12
    Ah, I was trying to do both equations, in two different columns. Maybe that added to the confusion. I didn't realize until now how close the spacing was. The acceleration for the helicopter going up is positive 15m/s^2,correct? Thanks for explaining!

    My new simplification is posted below. I think it looks good, but I am still unsure of the next step. Do I use the quadratic formula? Thanks again!

    x=x0+v0t+1/2at^2 x=x0+v0t+1/2at^2

    30 m=60 m+v0t+1/2 (15m/s^2)t^2 30 m=60m+v0t+1/2(-9.8 m/s^2)t^2

    30m=60m+v0t+(7.5m/s^2)t^2 30m=60m+v0t+(-4.9m/s^2)t^2

    -30m=v0t+(7.5m/s^2)t^2 -30m=v0t+(-4.9m/s^2)t^2
     
  14. Dec 6, 2013 #13
    For the helicopter:

    x=x0+v0t+1/2at^2
    30m=60m+v0t+1/2(15m/s^2)t^2
    -30m=v0t+(7.5m/s^2)t^2

    Thank you for explaining! When I posted, it "corrected" my spacing, which seemed to make reading the process slightly more complicated. Does that simplification look better? What's next? Should I use the quadratic formula?
     
  15. Dec 6, 2013 #14
    What is the initial speed of the helicopter? And initial position?
    Did'n you use the origin at ground level?
     
  16. Dec 6, 2013 #15
    Posted twice.
     
  17. Dec 6, 2013 #16
    The initial speed, as in the velocity? I'm not really sure. Before it launched, it would be zero. Oh, thank you for pointing out it started at the ground. I used the building as the starting position. I took 60 m to be the initial position. I was attempting to simplify by moving 60 m to the left, but I'm not sure it worked.

    Thanks!

    Is this better?

    x=x0+v0t+1/2at^2

    60 m=0m+v0t+1/2(15m/s^2)t^2

    60m=v0t+(7.5m/s^2)t^2

    That still looks like it could be a quadratic equation. Should I try to set it to zero?
     
  18. Dec 6, 2013 #17
    I think you are mixing up the two now.
    Helicopter: starts from ground (xo=0), no initial speed (vo=0). Acceleration is 15m/s^2 upward.
    x1(t)=1/2 a*t^2. At the collision point x1=30 m.

    Rock: starts from 60 m above ground (xo=60m) vith unknown initial speed (vo). Acceleration is g downward (about 10 m/s^2, minus sign)
    x2(t)=60m+vo*t-1/2g*t^2.

    At the collission point, x2=30 m.

    Try now. Yes, the second equation is quadratic in t. But you don't have to solve a quadratic. You need vo.
     
  19. Dec 6, 2013 #18
    Hmm, for the helicopter, like this, perhaps?

    x=x0+v0t+1/2at^2

    30m=0+0+1/2(15m/s^2)t^2

    30m=7.5m/s^2(t^2)

    t^2=4

    t=2 s

    This looks fairly reasonable. The meters cancel out nicely, and so do the seconds^2. Is this the correct time for the helicopter?
     
  20. Dec 6, 2013 #19
    Yes, it look OK.
    Now write equation for the falling rock and put in this time (2s).
     
  21. Dec 6, 2013 #20
    Arrrr. Sorry, posted twice again.
     
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