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Homework Help: Ball in Flight

  1. Apr 9, 2005 #1
    A ball is thrown into the air from ground level. After a time t = 1 s, the ball has traveled to a position x1 = 28 m to the right of and y1 = 13 m up from where it was thrown (at this time, the x and y components of the ball's velocity are still positive). The axes show the x and y directions to be considered positive.

    a) What was the x component of the initial velocity of the ball?
    v0x = m/s *

    b) What was the y component of the initial velocity of the ball?
    v0y = m/s *
    15.97 NO

    HELP: Look in your book for an equation.
    HELP: You need an equation relating initial velocity, the change in position, acceleration, and time.

    c) What was the initial speed of the throw?
    v0 = m/s *

    HELP: Use the answers from parts (a) and (b).

    d) What was the initial angle of the throw relative to the horizontal? Please enter your answer in degrees.
    q0 = ° *

    e) What is the height of the ball at the top of its path?
    h = m
    8.1 NO

    I need help with part e, that's it

  2. jcsd
  3. Apr 9, 2005 #2

    Andrew Mason

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    Use energy. What is the energy of the ball when thrown? When that energy is all converted into potential energy, what will its height be?

  4. Apr 9, 2005 #3
    Not sure if using energy will work as at the top of the parabolic path horizontal component velocity would be the same as at the point of being launched so not all amount of energy at the start would be converted to P.E or am I wrong?

    Try using this equation,

    v^2 = u^2 + 2as

    where u is vertical component velocity at launch, v is final vertical velocity, a is g and s distance travelled in the y axis.
  5. Apr 9, 2005 #4
    Well, I did this:

    v - final velocity = 0

    a = -g = - 9.81 m/s^2

    so, -u^2/-2g

    Can we use 'u' from part c.

    do you think this would work.
  6. Apr 10, 2005 #5

    Andrew Mason

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    As Al pointed out, you can only use vertical velocity or its vertical kinetic energy. Its maximum height is given by:

    (1) [tex]mgh = \frac{1}{2}mv_{0y}^2[/tex]

    You can also use:

    (2) [tex]y = v_{y0}t - \frac{1}{2}gt^2[/tex] and

    (3) [tex]v_{y} = v_{y0} - gt[/tex]

    and find t when v_y = 0 (ie. [itex]t = v_{y0} /g[/itex]. Substituting that value for t into (2) gives you maximum y. But you can see that it works out to the same as h in (1).

  7. Apr 10, 2005 #6

    You could use Andrew's method or if you'd prefer to use back the equation I wrote earlier, the u, has to be the vertical component of the initial velocity. i.e, you cannot use from (c). Use answer from (b), the y component velocity.
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