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More physics Help Please

  1. Nov 28, 2004 #1
    More physics.... Help Please!!

    A ball player hits a home run, and the baseball just clears a wall 18.5 m high located 140.0 m from home plate. The ball is hit at an angle of 32° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.0 m above the ground.

    a. What is the initial speed of the ball? Solve for time in the x-direction and substitute it in the formula for the vertical position.
    Incorrect. Tries 1/15

    b. How much time does it take for the ball to reach the wall? Tries 0/15

    c. Find the velocity components and the speed of the ball when it reaches the wall.
    Vy,f
    Vx,f
    Vf

    -----------------
    I have no idea how to start :yuck:
     
  2. jcsd
  3. Nov 28, 2004 #2
    Well let's start with what we know --

    We seem to be told to assume that the 18.5 m wall is the apex of the ball's height and the ball started at 1 m so it travelled 17.5 m up before its vertical vel went to zero (vertical velocity is zero when the ball stops rising and starts falling). Velocity after time t is initial vertical velocity (Vy here I think) - gt -- that is:

    V(17.5 m) = Vy - gt = Vy - 9.8t

    If we set that to 0 we get that

    Vy = 9.8t -------- first step

    We also know that the distance up (17.5 m) is as follows

    17.5 = Vyt - gt^2/2 --- But note that we have an expression for Vy just above so

    17.5 = 9.8t^2 - 9.8t^2/2 = 4.9t^2

    t = SQRT(17.5/4.9) = 1.89 sec

    So Vy when the ball left the bat was:

    9.8 * 1.89 = 18.522 m/sec

    I did this straight on the computer and I have not checked whether I made any errors -- so check it -- the logic should be right.

    Why don't I let you work the rest.
     
    Last edited: Nov 28, 2004
  4. Nov 28, 2004 #3
    Thank you for your help, but my online homework says that is the wrong answer. So if someone could still help me, I'd appreciate it.
     
  5. Nov 29, 2004 #4
    OK - one more try; I appear to have been wrong in assuming that 18.5 m was the apex of the ball's flight. Using the hint from part a.,
    where t is the time at which the ball reached the fence
    travel in x direction:

    140 m = Vx*t so
    t=140/Vx

    travel in y direction:
    17.5 = Vy*t - 9.8*t^2/2

    17.5 = (Vy/Vx)*140 - 0.5*9.8*19600/Vx^2 (equation 1)

    I think we assume tan(32)= Vy/Vx

    i.e., 0.6249 = Vy/Vx or Vy = 0.62489*Vx

    Using this to solve equation 1 above yields Vx= 37.045 m/sec so
    Vy = .6249*37.045 = 23.14 m/sec.
    For "initial speed" they probably want the velocity at 32 deg from the horiz so that would be Vy/cos(32) or Vx/sin(32)
    Well, hope this one was right
    I'll leave b) for you
     
    Last edited: Nov 29, 2004
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