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1d kinematics help - ball thrown vertically upwards

  1. Jul 26, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball is launched vertically upward from ground level with initial speed of 20 m/s.
    1. How long is the ball in air?
    2. What is the greatest height reached by the ball.
    3. How many seconds after launch is the ball 15m above the release point.
    (Air resistance is negligible)

    The answers from the back of the book are:
    1. 4.1 s
    2. 20 m
    3. 0.99 s and 3.1 s


    2. Relevant equations
    Kinematic equations for constant acceleration:
    1. vx = v0x + axt
    2. vav x = 0.5(v0x + vx)
    3. [tex]\Delta[/tex]x = v0xt + 0.5axt2
    4. [tex]\Delta[/tex]x = 0.5(v0x + vx)t

    3. The attempt at a solution

    Part 1
    To find how long the ball is in the air - I need to find t (or 2t depending how you look at it)
    initial velocity is 20m/s and I am assuming final velocity to be zero.
    a is 9.8

    So I used equation 1 above:
    So 0 = 20 + 9.8t => t=2.0408 (ignoring the sign)
    Hence 2t = 4.0816 (the total time ball is in the air)
    Is this correct, did the textbook round it up?

    Part 2
    What is the greatest height reached by the ball?
    I use equation 4 above to find delta x:
    [tex]\Delta[/tex]x = 0.5(20+0)2.0408
    I get 20.408 m
    Again is this correct, did the textbook round it up?

    Part 3
    How many seconds after launch is the ball 15m above the release point?
    I am assuming we are given [tex]\Delta[/tex]x as 15m?
    I used equations 3 and 4 here but my answer is nowhere close.

    Can someone please help me. Thanks in advance
    Okay I don't know how to proceed here.
     
  2. jcsd
  3. Jul 26, 2009 #2
    Part one and two look correct.

    You're looking for the time when x >= 15m right?

    Well looking at equation 3 it's a parabola with x(t) = vt +0.5at^2. Just looking at the graph the ball is at x=15m twice in it's flight, once while it's going up and once on the way back down.

    Does this help?
     
  4. Jul 26, 2009 #3
    Thanks for replying.
    Yes, the ball would be at 15m twice in flight.
    Hence from equation 3 is tried:

    4.9t2 + 20t - 15 = 0
    The solution is -4.728 and 0.647 none of which make sense to me or are even close to the answer.
     
  5. Jul 26, 2009 #4
    The required equation is:
    4.9t2 - 20t + 15 = 0

    because

    h = ut - 1/2gt2

    since the ball is launched vertically upward g become negative
     
  6. Jul 26, 2009 #5
    Ahh...Thanks so much.
    Stupid signs...
     
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