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Projectile motion of a ball thrown

  1. Sep 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Julie Throws a ball to her friend sarah. The ball leaves julies hand a distance 1.5m above the ground with an initial speed of 23m/s at an angle of 26° wrt the horizontal. Sarah catches the ball 1.5m above the ground. (a) what is the horizontal component of the balls velocity when it leaves julie's hand (b) what is the vertical component of the balls velocity when it leaves julie's hand
    (c) what is the maximum height the ball goes above the ground? (d) what is the distance between the two girls? (e) After catching the ball sarah throws it back to julie. The ball leaves sarah's hand a distance of 1.5m above the ground, and it is moving with a speed of 25m/s when it reaches a maximum height of 9m above the ground. What is the speed of the ball when it reaches sarahs hand? (f) how high above the ground will the ball be when it gets to julie?
    2. Relevant equations

    xf = xi + v0t +.5at^2
    vf^2 = vi^2 + 2a(x - x0)
    v = v0 + at

    3. The attempt at a solution
    I got to the last part but I want to check my work over the previous parts, there are no answers in this book.

    for (a) and (b) I have vx = 23cos(26°) and vy = 23sin(26°) for (c) i want to know the final height over some interval that is not the entire distance, I had to calculate the total time that the ball was in the air which I got using yf = yi +v0yt +.5at2 with yf = yi = 1.5m

    I got 0 = 10.08253638(T)m/s - 4.905m/s2t2m/s2
    and the total time as 2.055562922 seconds
    I usually can divide a lot of these total distances and total times by 2 because of the symmetry of the parabolic path the ball travels. So I went ahead and found t/2 = 1.027781486
    so yf(t/2) should give me the maximum height of the ball
    yf = 1.5m + 10.08253638(1.1.02778148656) + .5(-9.81m/s^2)*(1.027781486)^2
    yf = 7.43m
    okay
    the next question asks me to find the total distance
    xf = xi + v0x + .5at2
    where xi = 0m + v0x =
    23cos(26°) t = 2.05...seconds and for "a" I still used the adtg as 9.81m/s^2
    so xf = 63.22m about 180 feet, this one is really long(probably because ax = 0 there is constant velocity in the x, maybe the real distance is 41.34m(got this by just plugging in ax = 0)
    okay
    (e) calculate the initial speed of the ball when it leaves sarah's hand
    okay I need vxi and vyi to get at this initial speed. I tried using vfx = vix + at but vfx = vix so I actually can't use the work I wrote down, looking at it again I must concede I did that part incorrectly, which leaves me at (e) i am not sure how to do this part. maybe I could start by finding vx at the total distance or something?
     
    Last edited: Sep 20, 2013
  2. jcsd
  3. Sep 20, 2013 #2
    or perhaps something along the lines of a substitution?
    vf2 = vi2 + at
    hmm maybe I can eliminate one of the two unknowns in these expressions, vfy or t by doing this? If it's possible i'm not seeing it.
     
  4. Sep 21, 2013 #3

    haruspex

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    Another way is to put vertical speed = 0 at max height and use vf = vi + at.
    A good habit to get into is working entirely symbolically until the final step. That makes it much easier for others to follow what you are doing, avoids accumulation of rounding errors, and makes it easier to spot mistakes. And you certainly don't need to be handling so many significant figures. 3 or 4 is plenty.
    A couple of problems there. You are missing a factor t in the v0 term, and gravity does not generaly act horizontally.
     
  5. Sep 21, 2013 #4
    okay, so a_x = 0
    that should make the expression
    xf= xi+ v0x*t
    xf = 0m + 23cos(26)m/s*(2.05seconds)
    xf = 42.37m
    I am assuming for the next part we should be using the result from part d) that I calculated above
    so is this part asking me, "when xf = 42.37m what is yf"? that seems reasonable to me, i'm just drawing a blank when it comes to trying to come up with an expression that relates the two, can I set xf = yf and try to solve for something there?
     
    Last edited: Sep 21, 2013
  6. Sep 22, 2013 #5

    haruspex

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    Yes, you can use the distance you calculated between the girls, but everything else is different.
    You are told the speed of the ball when at max height. What are the horizontal and vertical components of that speed?
     
  7. Feb 5, 2014 #6
    Hey guys, I'm now taking Fundamentals of Physics 1 and I've encountered this same problem. I'm having difficulty with problem c) find the maximum height.
    What I did was I used Vy = Voy + ayt, and I set Vy to 0 and solved for t to find the time at which the vertical velocity was 0.
    I then just plugged that value of t into the vertical position equation, but I got a wrong answer. Could anyone please explain to me why i can't do it that way? Everywhere I look suggests that I use a different method with a different equation, namely the one in the quote I used.

    Could someone please help me with this? Thank you so much.
     
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