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Homework Help: 2 variables!

  1. Sep 26, 2006 #1
    I have to say, Physics has been going well for me. But he assigned a problem in preparation for a test. Well needless to say, I am not sure where to begin! I was hoping somebody could at least point me in the right direction.

    A baseball is hit for a home run, it travels 138m. It just clears a 6.5m wall when it lands, where it is caught by a lucky fan. So the ball essentially starts off at point (0,0), and ends at (138,6.5).
    It is hit at a 40 degree angle. Calculate initial velocity and total flight time.

    I know it must be possible. Because only at one velocity at 40degrees will the ball actually go 138m. Like if it was hit at 1m/s, obviously it would not go the full 138. It's finding this balance point. My thought was to first find velocity, THEN find time. If only i knew how

    I've got a few ideas brewing upstairs, but ugh.
  2. jcsd
  3. Sep 26, 2006 #2


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    Step 1: present your work, and we will be glad to help you. :smile: (Hint: write down the equations of velocity and position of the ball.)
  4. Sep 26, 2006 #3
    I am stuck because, well I want to break it down into two vectors. So say I use X1=X0+V0t+1/2at^2

    For one vector, the j vector because it has acceleration. I don't know time or initial velocity, I don't understand how I find what I need to find.
    So 6.5=0 + Vot + -4.905t^2

    I would think I need initial velocity to calculate this.
  5. Sep 26, 2006 #4
  6. Sep 27, 2006 #5


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    Well, you just have to solve a system of two equations with two unknowns, which are the initial velocity and the total 'flight time'. The equation of motion for the x-direction is x(t) = v0*cosA*t , and for the y-position y(t) = v0*sinA*t - 1/2*g*t^2. (A is, of course, the angle of the initial velocity.) Now, since you know the x and y coordinate at the end of the 'flight', just plug them in and solve the system of equations.
  7. Sep 27, 2006 #6


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    The parabolic equation for a projectile is of the form

    [tex]y = Ax - Bx^2[/tex]


    [tex]A = \tan(\theta _o)[/tex]


    [tex]B = \frac{g}{2(v_o \cos(\theta _o))^2}[/tex]

    Another solution:
    Shoot the monkey - d is the distance the poor creature drops through before encountering the projectile. h the height at which the monkey is shot: 6.5 meter. Then

    [tex]\tan(\theta _o) = \frac{d+h}{138}[/tex]

    the rest is easy .... or is it? I do get the feeling that this is the appoach your teacher expects from you.
    Last edited: Sep 27, 2006
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