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A ball is thrown up at 20 m/s at an agle of 15 degrees

  1. Aug 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown upwards at 20 m/s at an angle of 15 degrees to ground.
    How long until the ball touches the ground?

    Givens for vertical:
    Acceleration = -9.8m/s2
    v1= 5.2 m/s (20SIN15)
    v2=0
    displacement=0
    t=?

    Givens for horizontal
    Acceleration= 19.3 m/s (20COS15)
    Displacement= ? (perhaps work back from displacement on horizontal to get the time?)
    t=?

    2. Relevant equations

    DeltaD= (V1)(DeltaT) + (.5)(A)(DeltaT)
    Answer should be 0 OR 1.1s for T ( I don't understand how you are to get two answers)
    3. The attempt at a solution
    0=(5.2)(detlaT) +(0.5)(-9.8)(deltaT)

    ?? ?? ??
    The textbook suggests to use this equation and not the other 4.

    Also I am not sure if displacement is zero for the horzontal given that V=D/T


    Stumped for hours.
    Nobody I can ask for help
     
  2. jcsd
  3. Aug 21, 2015 #2
    To get the answer the problem requires, I suggest that you just focus on the vertical velocity, with the gravitational acceleration as a constant.
     
  4. Aug 21, 2015 #3
    I get that. I don't understand why the book suggests there are two answers 0 and 1.1
    I am missing something basic I think.
     
  5. Aug 21, 2015 #4
    That's because at the first ##(t=0)## the ball is also on the ground, so if you use the displacement to solve the quadratic equation, you will get two answer, of course one of which is trivial.
     
  6. Aug 21, 2015 #5
    Yeah, it's an issue with my math. I have not taken math in high school since the original September 11th, so you can guess my age....
    What is the operation called apart from quadratic equation. I have to figure this stuff out from square one. Thank you for the help.
     
  7. Aug 21, 2015 #6
    If you set the time ##t## and the displacement ##d,## you can get
    $$v_{0,ver}t+\frac{1}{2}(-g)t^2=0$$
    and one of the solutions is ##0.##
    The way I suggested above:
    $$-v_{0,ver}=v_{0,ver}+(-g)t,$$
    which can be intuitively get by
    $$t=\frac{2v_{0,ver}}{g}$$
    by the process of constant acceleration.
     
  8. Aug 22, 2015 #7
    When solving projectile motion problems, it tends to be in your best interest to stick with the variable you have the most information on. Projectile motion problems tend to require you to first find t, which is the only variable shared by both x and y, as both are functions of t. If you, for instance, were asked to find the displacement in x instead of finding t, then you would have had to find t first and then use kinematics equations on the x plane instead of the y plane.
     
  9. Aug 22, 2015 #8
    A simple way to solve this problem is to find the initial vertical velocity of the ball. When the ball is at the top of its arc, its velocity is zero. Using kinematic equations, the time it took for it to reach the top can be found. Due to the conservation of energy, the time it took to get to the top will be equal to the time it takes to hit the ground.
     
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