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Projectile Motion and time

  1. Dec 8, 2004 #1
    A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 105 m/s at an angle of 37 degrees with the horizontal. What is the time it takes the projectile to hit the ground? How far from the base of the cliff does it land?

    Now my major problem is that I don't know how to set up the motion equation using the 37 degrees.....help!! i've done the problem before and I turned it in and it was wrong.
     
  2. jcsd
  3. Dec 8, 2004 #2

    Diane_

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    Split the initial velocity into horizontal and vertical components. Use the vertical component to find the time of flight. Let's assume that you define up as positive and down as negative: you'll know the initial velocity, the acceleration, and the final position (-125 m). From that, you can find the time.

    Once you have that, use the horizontal component to find the distance: you'll have the initial velocity, the acceleration (which should be obvious), and the time.
     
  4. Dec 8, 2004 #3
    figure out the x and y components then then put it into the equation d=v1t+1/2at^2
     
  5. Dec 8, 2004 #4
    but how do I find the vector components,...im so confused!
     
  6. Dec 8, 2004 #5

    Diane_

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    Use right-triangle trigonometry. Sketch a right triangle with the hypotenuse as the initial velocity - the 105 m/s. Remember, then, that the sine of an angle is the opposite leg divided by the hypotenuse and the cosine is the adjacent leg over the hypotenuse.
     
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