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Homework Help: Projectile motion equation

  1. Aug 23, 2011 #1
    1. this is straight out of PHYSICS 8e by cutnell and johnson

    chapter 3 page 70

    there are two solutions to this equation. one is given by (14 m/s + 1/2 (-9.8 m/s ^2)t = 0 or t = 2.9s

    what i cant seem to work out is how did they get t = 2.9s from those calculations

    2. y = Voyt + 1/2 ayt^2

    3. i just cant seem to calculate the final answer .... do i need to rearrange the equation more ??
  2. jcsd
  3. Aug 23, 2011 #2


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    We can't help you if we don't know the question!
  4. Aug 23, 2011 #3
    given the initial velocity, it is acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is Voy = +14 m/s, therefore we have y = 0m ay = -9.8m/s^2 V0y = +14m/s t = ?

    The time of flight can be determined by the equation y = voyt + 1/2ayt^2
  5. Aug 23, 2011 #4


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    Correct so far. Since the equation you quote is quadratic, it has two solutions. The first is obviously t=0 (it starts off at ground level). This leaves

    [tex]0 = v_0 + \frac{1}{2}at[/tex]


    [tex]t = -\frac{2 v_0}{a}[/tex]

    Do you follow?
  6. Aug 23, 2011 #5
    i understand a quadratic has 2 solutions but i cant seem to see how that equation is a quadratic
  7. Aug 23, 2011 #6


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    [tex]y = v_0 t + \frac{1}{2}at^{\color{red}2}[/tex]

    The power "2" makes it a quadratic by definition!
  8. Aug 23, 2011 #7
    Anything thrown upwards (at least on the earth) will have a parabolic shape - which is a quadratic.
    Last edited: Aug 23, 2011
  9. Aug 23, 2011 #8


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    You could make use of the fact that the projectile come down just as fast as it went up, so final velocity is -14 m/s

    That way you an use the non-quadratic formula v = u + at [not your symbols but a common set]
  10. Aug 23, 2011 #9
    You don't need to use the quadratic equation.
    I offer a more pictorial method for this problem.

    There are two important times during the parabola trajectory, tA and tB.
    tA is when the object is at the peak of the parabola.
    tB is when the object lands.

    tB = 2tA

    **since acceleration caused by gravity is constant
    Ay = (Vf - Vi) / tA

    Ay = -g

    -gtA + Vi = Vf

    **Vf=0 because Ay is ZERO at the apex of the parabola

    tA = -Vi/-g =1.42s

    2tA = 2.9s = tB

    Go to this link http://books.google.com/books?id=6u...resnum=1&ved=0CCEQ6AEwAA#v=onepage&q&f=false"

    Go to page 80. You will see 2 graphs of parabola trajectory for projectile motion. Look at figure 4.9.
    Last edited by a moderator: Apr 26, 2017
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