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Dynamics: Normal and Tangential projectile motion problem

  1. Jun 23, 2014 #1
    1. The problem statement, all variables and given/known data

    A canon fires a projectile at a speed of 100 m/s at angle of 20 degrees. (a) what is the radius of curvature of the projectile at its highest altitude? (b) What is the radius of curvature of the path .5 seconds after firing?

    2. Relevant equations

    Velocity vector = s'ur

    Acceleration vector = v' =
    atut + anun

    where,

    at= v'= dv/dt

    &

    an= [v^(2)]/p


    3. The attempt at a solution

    Wrote out the given which was,

    Vo = 100 m/s at an angle 20 degrees above the horizontal x+ axis

    Vox= Vo*cos(20)

    Voy= Vo*sin(20)

    ax = 0 = at = 0

    ay= g or -9.81 acting in the negative j direction.

    (a) an = -9.81 ==> -9.81 = [v^2]/p ==> p = [100^2]/9.81 = 1019.37 meters
     
  2. jcsd
  3. Jun 24, 2014 #2

    BruceW

    User Avatar
    Homework Helper

    hey! welcome to physicsforums :)

    Uh, your answer for part a is almost right. But v is not 100m/s. Because 100m/s is the initial velocity. So what velocity does the object have when at the highest altitude?
     
  4. Jun 24, 2014 #3
    Thanks for the welcome and help!

    Well at the highest altitude I guess the y component of the velocity is zero, so I would only have to calculate the x component of velocity. Therefore, an= [v^2]/p is actually

    -9.81=[(100m/s*cos 20)^2]/p .. solve for row and p= 900.12 meters? Would I use the equations (Vosin theta)t and (Vocos theta)t at time .5 seconds? the are vectors so I could find my speed by Pythagorean theorem.
     
  5. Jun 24, 2014 #4

    BruceW

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    Homework Helper

    yes, I agree with your answer of 900.12m for part a. And for part b, yes, that sounds like a good plan. find the vector velocity at that time, and then you can find both an and |v|. (remember that an is not the total acceleration, but the acceleration in the direction perpendicular to the velocity).
     
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