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A rock is thrown at an angle of 17 degrees above the horizontal at 28.5 m/s, from the

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A rock is thrown at an angle of 17 degrees above the horizontal at 28.5 m/s, from the ground.

    a)How high will the rock rise above the ground?
    b)What is the horizontal speed of the rock at the top of its flight?
    c) How long will it take the rock to reach a vertical speed of zero m/s?


    3. The attempt at a solution
    First I found out what the x and y components are. x=27.25 m/s and y=8.35 m/s. Then I found the time w/ vf=vi+at. (27.25=9.8t)

    a)I used the forumla d=vit+1/2at^2 and it came out to 37.87 m
    =1/2(9.8)(2.78)^2
    b) I'm guessing I would use the formula v=d/t. So the max height is 37.87 and the time is 2.78(1/2 the time of the rock's total air time). and I got the answer 13.62 m/s
    c) I wasnt really sure on this one. Is it 2.78s, when the ball is at the top of its flight?

    Any advice/help would be great. I want to make sure I'm doing this correctly. Thanks! :)
     
  2. jcsd
  3. Feb 7, 2009 #2
    Re: A rock is thrown at an angle of 17 degrees above the horizontal at 28.5 m/s, from

    The horizontal velocity never changes
     
  4. Feb 7, 2009 #3
    Re: A rock is thrown at an angle of 17 degrees above the horizontal at 28.5 m/s, from

    This equation will be helpful
    The vertical position with respect to time is given by
    y(t) = y_0 + (V_0)t-4.9t^2
    V_0 in the y direction is V_0sin(theta)
    so 28.5sin(17)
    where y'(t) = 0, the projectile has reached it's max height. Put that t into the y(t) to find the maximum height. This should also be the answer for c)
    velocity in the x direction is V_0cos(theta)
    so 28.5cos(17). As casedogg said, it doesn't change.
     
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