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Physics Projectile Launched at Angle. HELP, TEST TODAY!

  1. Oct 4, 2011 #1
    Physics Projectile Launched at Angle. HELP, TEST TODAY IN 2 HOURS!

    1. The problem statement, all variables and given/known data
    An object is launched from the top of a building
    with an initial velocity of 15 m/s [32 degrees]. If the
    building is 65.0 m high, how far from the base
    of the building will the object land?


    2. Relevant equations
    dx = vix (t)
    dy=viy(t) + 1/2a (t)^2


    3. The attempt at a solution
    First I drew a diagram. Then set up the x and y table:
    The variables I know; vix= 12.72 (which I got from vicos(theta), viy= 7.948, ay=-9.81, ax=0.

    Then I solved the time for the 32 degrees. Then I solved the time for the other part and then I solved for dx, I got 66.9 m.

    The answer in the book says its 58 m.

    Can't figure out how to do this, help please!!!
     
    Last edited: Oct 4, 2011
  2. jcsd
  3. Oct 4, 2011 #2

    gneill

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    Staff: Mentor

    Re: Physics Projectile Launched at Angle. HELP, TEST TODAY IN 2 HOURS!

    Can you show your work for these steps?
     
  4. Oct 4, 2011 #3
    Book answer looks good. I'm not quite sure what you did, though.

    First step, which you've done correctly, is to find independant velocity of each component. As you've identitfied, the y velocity is 15sin32, while the x is 15cos32 (7.9 and 12.7).

    So, model the y-position with this info.

    You know acceleration is due to gravity, and you know that the velocity is 7.9 m/s. You also know initial y position as 65 meters. This gives you a quadratic equation. You need to solve for t when the equation is equal to 0 (looks like you know this).

    When you have that t, you should build a x position formula. Since there is no acceleration and no initial position, it will just involve velocity along x. Solve this equation for the t value you find.

    Could you show what your y position equation looks like, and what your time is and how you got to it?
     
  5. Oct 4, 2011 #4
    Now I'm getting 46m. I do: 65= 7.9(t) + 1/2a (t)^2
    And I solve for t.

    Then I use the time I found and multiple that by the Vix. And then I get 46.228m, which is wrong.
     
  6. Oct 4, 2011 #5
    Your y equation ignores the fact that you are starting from a height of 65m.
     
  7. Oct 4, 2011 #6
    I fixed it, I did use 65m, just forgot to write that in.

    I am still getting 46m.

    65=7.9t + 1/2a (t)^2

    then I use this to solve for t: √2(d) divided by acceleration. Which gets me 3.64s.

    Then I do 3.64s * the horizontal velocity which is 12.72 = 46.3 m. Which is still wrong.
     
  8. Oct 4, 2011 #7
    I have no idea what kind of algebra you're doing.

    -4.9t^2 + 7.9t + 65 = 0

    Quadratic formula.
     
  9. Oct 4, 2011 #8

    gneill

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    Staff: Mentor

    There are some "relationship" problems in your equation. You want to make sure that it corresponds to what happens physically. To do that I find it helpful to put the 'after' stuff on one side and the 'before and during' stuff on the other.

    So when the object lands after its trip it will be at height y = 0. That goes on the left hand side. On the right, the object starts at initial height 65m, initial velocity 7.95m/s upwards, and accelerates at 9.81m/s2 downwards. So:

    0 = 65m + (7.95m/s)*t - (1/2)(9.81m/s2)*t2
     
  10. Oct 4, 2011 #9
    Hint: The time of flight is greater than what you calculate. The time of flight is made up of two components. One is the time for projo going up; the other is the time projo going down.

    So compute how high the projo goes above the building. Compute this time. Then compute the time for a free fall from maximum altitude above building roof to the ground. Add them together. You know the rest...
     
  11. Oct 4, 2011 #10

    gneill

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    Staff: Mentor

    While one could do it that way, it is not necessary to carve up the trajectory in to sections.

    The vertical trajectory equation, y = yo + vo*t - (1/2)*g*t2 applies to the whole trajectory and will yield the correct time no matter the initial velocity or elevations of the launch, apex, or landing.
     
  12. Oct 4, 2011 #11
    I'm well aware of that. But it can make it more understandable to the student if he breaks the flight into parts rather than merely plugging and chugging with equations that include additional terms. Students must understand the nitty gritty of problems, not just plug numbers into equations.
     
  13. Oct 4, 2011 #12
    The parabolic equation is intuitive. OP is just making small algebraic mistakes but understands the concept.
     
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