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The flying baseball

  1. Oct 28, 2005 #1
    I couldn't get my head around this problem... I'd like some input and some answers to aim towards please... I'm actually a pretty bright student, I just can't figure this out... maybe a blonde day lol? :smile:

    A baseball of mass 55g leaves the pitchers hand at a speed of 48m/s. THe horizontal distance to the batter is 20m. Assuming that air resistance is negligible, determine the following:

    a) the time for the ball to travel the first 10m
    b) the time for the ball to travel the second 10m
    c) the distance that the ball fell in the first 10m of horizontal travel
    d) the vertical velocity of the ball just before it reached the batter.

    If the ball missedthe bat and was taken by the catcher, whose glove moved backwards 30cm as he caught the ball, calculate the following:

    e) the change in velocity of the ball as it was caught
    f) the acceleration of the ball as it was caught
    g) the force exerted on the glove by the ball as it was caught
    h) the force exerted on the ball by the glove as it was caught
     
  2. jcsd
  3. Oct 28, 2005 #2
    i got the first half!

    hey shep!
    i solved the first half of ur quest....
    this is a projectile problem thats launched horizontally.....
    vx is given thats 48m/s and the totaly x is given =20m
    so first thin we cud solve for total time...x/vx(20/48=0.42s
    for first ten m....we will put x=10m
    10/48= 0.21 s....time taken ta travel first 10 m....
    as we know the total time is 0.42 s....so we will simple get the difference between the total time and time taken ta travel the first 10 m....to get the second halve's time.
    0.42-0.21=0.21s
    to get the vertical distance covered in the first 10 m....
    we know initial vy(vertical velocity has to be 0, and time taken in first 10 m is 0.21s....
    y= ut+.5at^2......y=0+.5(9.8)(0.0441)
    =0.22m......
    now ta get the final vertical velocity...we will use the simple equation..vyf=vyi+at
    v= 0+9.8 (0.42)
    =4.1m/s
    i guess i will take a quick nap...and then try ta solve the next part.....ha a long de!
    b4 that if u get the hang of this question....do try n post the answer!
    take care.....
     
  4. Oct 29, 2005 #3
    Thanks for your working so far :smile: you make a lot more sense than my physics teacher, I have to tell you that! I had this strange idea that W=Fscos(theta) would be of use... and W=Fssin(theta) also...
     
    Last edited: Oct 29, 2005
  5. Oct 29, 2005 #4

    daniel_i_l

    User Avatar
    Gold Member

    second half:
    I assume that after the 30 cm the ball stopped. so the change in velocity would be the speed of the ball just before it got to the catcher.
    f) If you know the final and initial speeds and the distance that it took to go from one to the other than to find the acceleration you can use:
    Vf^2 = Vi^2 + 2*a*x tell me if you want to see the derivation.
    g) with the acceleration you can find the time, and if you find the impulse you can get the force.
    h) Newtons 3rd law.
     
  6. Oct 29, 2005 #5
    Thanks for all this help :smile: ... I think I'm going to stick around on this board and help other ppl out lol
     
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