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Kinematics - Finding velocity of a baseball

  1. Feb 23, 2005 #1
    Hi I have a question
    A baseball player hits a baseball that lands in stands 24 meters above the playing field.The ball land with velocity of 50 meters per second at angle of 35 degree below the horizontal.

    a.If a player hits the ball 1 meter above the playing field,what was the velocity of the ball upon leaving the bat?
    b.What was the horizontal distance traveled by ball
    c.how long was the ball in air

    I have done like this:

    let r be angle in degrees

    t=2usinr/g

    time=?
    velocity=50m/s
    gravity=10

    time= 2*50*sin35/10=5.74s
    velocty of ball at 1m will be:

    y=vsinrt-0.5gt
    1=usin35*5.74-0.5*10*5.74*5.74
    u=50.61m/s


    b.horizontal distance traveled by ball

    distance/time=velocity
    then
    distance/time=ucosr
    distance=50cos35*5.74
    distance=235m

    c.Time of the ball in air.

    t=2usinr/g

    time=?
    unitial velocity=50m/s
    gravity=10

    time= 2*50*sin35/10=5.74s

    Any idea about this question?
     
  2. jcsd
  3. Feb 23, 2005 #2
    Let the horizontal and vertical components of the initial velocity be [tex]u_1, u_2[/tex]

    Apply [tex] v^2 = u^2 + 2fS[/tex] in vertical and horizonatal directions to find u1 and u2.

    Note that the vertical distance travelled is 24 -1 = 23 m.

    So inital velocity = [tex]\sqrt {(u_1^2 + u_2 ^2)}[/tex]

    (2). Apply [tex] v = u + ft [/tex] vertically to find the time of travell

    (3) Apply [tex] s = ut + \frac{1}{2} f t^2 [/tex] horizontally to find the distance travelled.
     
  4. Feb 24, 2005 #3
    Ooo is fine but my doubt is about the angle 35 degees u didn't mention it
     
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