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Projectile motion

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A projectile is launched from the top of a cliff with an initial velocity of 20m/s at 30deg. above the horizontal. It hits the ground 3.0s later. Given projectile range is 52m and height is 14.1m, how do i calculate it velocity(magnitude and direction) 0.1s before it hits the ground?


    2. Relevant equations
    d=vit+1/2at^2
    vf=vi+at


    3. The attempt at a solution
    I already calculated the range and the height, but i don't know how to do the velocity before it hits the ground based on those figures in the question. Please hints? Do i have to draw another triangle for the landing and use a new value for time or something?
     
  2. jcsd
  3. Oct 6, 2009 #2

    Delphi51

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    Can't you just use y = Vi*t + .5*a*t^2 ?
    Vi is 20*sin(30) so you know everything.
     
  4. Oct 6, 2009 #3
    Well, they got 25.3m/s at 47deg below the horizontal...so...i'm not sure how they got there...any hints please? thanks.
     
  5. Oct 6, 2009 #4

    Delphi51

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    Sorry, Irish - I glossed over the problem and didn't see "magnitude and direction" and of course you need velocity formulas, not the distance one. The thing is the velocities for all times are given by
    Vx = 20*cos(30) and Vy = Vi + at = 20*sin(30) - gt
    so you just evaluate at the time you want. Then combine the two vectors - triangle, pythagorean theorem and a tan calc for the angle.
     
  6. Oct 7, 2009 #5
    I got a triangle with sides 10m/s vertical, 17.3m/s horizontal, and 20m/s hypotenuse. I"m lost now .....
     
  7. Oct 7, 2009 #6

    Delphi51

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    Vy = Vi + at = 20*sin(30) - gt at time 2.9 is -18.4
    so you have 17.3 to the right and 18.4 down.
    You want the angle with horizontal, the angle in the triangle adjacent to the 17.3 side. Just right tan of that angle = opposite/adjacent.
    Do inverse tan of both sides and you have your angle.
     
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