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Projectile Motion HELP

  1. Dec 9, 2007 #1
    A baseball leaves a bat at a rate of 50 m/s. The ball reaches its highest point after 4.42 seconds.
    (a) What is the angle with the horizontal at which the ball leaves the bat?
    (b) What is the shortest distance between the starting point and the highest point of the ball?

    Attempt:
    Vfy = Viy + a [tex]\times[/tex] [tex]\Delta[/tex]t
    :confused:
     
  2. jcsd
  3. Dec 9, 2007 #2
    This is a fairly basic projectile motion question, so you might want to brush up the concepts again.
    Yes, use vy=uy+ayt
    So the question here is what is vy at the topmost point. After you find uy, you can easily find the angle of projection using concepts of vectors.
    The second part follows easily once you have the horizontal component of the velocity.
     
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