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Problem with projectile velocity

  1. Oct 10, 2007 #1
    A ball is thrown leftward from the left edge of the roof of a building at height h above the ground. The ball hits the ground 1.50s later, at distance d=25.0m from the building and at angle theta = 60 degrees with the horizontal.

    A) Find h.

    What are the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown?




    I figured that initial velocity in X direction is 16.67 m/s (25 meters/1.5 seconds). What is really giving me trouble is finding the magnitude of the velocity. I can't figure out how to find the Y component of the velocity to get it. I'm not even sure what B and C want, but I figure i'll cross that bridge after I get A handled.
     
  2. jcsd
  3. Oct 10, 2007 #2

    learningphysics

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    Use the angle given 60 degrees, to find the height.
     
  4. Oct 10, 2007 #3
    Could you be more specific please? I don't know how to do that.
     
  5. Oct 10, 2007 #4
    If I am thinking of this right then your angle is opposite of your horizontal so you use tan(60)x25.0 m to find the horizontal.
     
  6. Oct 10, 2007 #5

    learningphysics

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    yes, tan(60)*25 is the height h.
     
  7. Oct 11, 2007 #6
    That is what I thought originally, too. Problem is the 25 is not the hypotenuse of the triangle that is formed, so (tan 60) * 25 yields an incorrect answer. That gives 43.3, but the back of the book says it is 32.3
     
  8. Oct 11, 2007 #7

    learningphysics

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    Ah... looks like I misunderstood the problem. The object is thrown at an angle, so it has an initial vertical velocity... it hits the ground with an angle of 60... that means that (vertical speed)/(horizontal speed) is tan(60) when the object hits.

    ie it hits with a vertical velocity of - 16.67tan(60) = -28.87m/s

    using this you can get the height, and it comes out to 32.3m.
     
  9. Oct 11, 2007 #8
    You're a lifesaver, thank you!
     
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