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Projectile motion

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A small ball is projected with a speed of 10 m/s at an angle of elevation of p from the top of a vertical pole whose height is 15 m from the level ground. The ball hits the ground at a point whose horizontal distance from the foot of the pole is 20 m . Find the value of p . gravity=10 m/s^2

    2. Relevant equations



    3. The attempt at a solution

    Consider horizontal motion , s=Vx t

    20=10 (cos p) t

    t=2/(cos p) --1

    Consider vertical motion,

    15=10 (sin p)(2/(cos p))+1/2 (10)(2\(cos p))^2

    15 cos^2 p -20 sin p cos p -20 =0

    3 cos 2p - 4 sin 2p=1

    5 cos (2p+53.13)=1

    2p+53.13=78.46

    therefore , p=12.67 degrees

    Am i correct ?
     
  2. jcsd
  3. Jun 5, 2010 #2

    rl.bhat

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    Homework Helper

    15=10 (sin p)(2/(cos p))-1/2 (10)(2\(cos p))^2

    Till the above step you are right except one sign. Next

    15 = 20tan(p) - 20sec^2(p)

    Put sec^2(p) = 1 + tan^2(p) and solve for tan(p)
     
  4. Jun 5, 2010 #3
    Thanks but isn't the ball being projected downwards which is in the path of the gravity pull ?
     
  5. Jun 5, 2010 #4

    rl.bhat

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    Homework Helper

    Projected at an angle of elevation means up against the gravitational pull.
     
  6. Jun 5, 2010 #5
    Be careful with your signs in your equation for the vertical position. We can use, since the acceleration due to gravity is approximately constant:

    [tex]\Delta y \equiv y_{f}-y_{0}=v_{y}t+\frac{1}{2}at^{2}[/tex]

    If we define our coordinate system so that up is positive and down, negative then we are given

    [tex]y_{0}=+15, v_{y}=+10sin(p), a=-g=-9.8[/tex]

    Since it starts out above the ground and is shot with an initial velocity up and gravity pulls it down.
     
  7. Jun 5, 2010 #6
    got it , thanks
     
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