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Projectile motion problem help

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data A home run is hit in such a way that the baseball just clears a wall 21m high, located 130m from homeplate. The ball is hit at an angle of 35 to the horzontal.Find the inital speed of the ball. Assume the ball is hit with a height of 1m above the ground



    2. Relevant equations



    3. The attempt at a solution[/b I don';t understand, how can I solve for a equation with two unknown variables. If I could only solve for max height I could solve this problems any tips. I've been trying to find a time independent equation for horizontal ranges to solve for initial speed. But that didn't work.
     
  2. jcsd
  3. Oct 12, 2009 #2

    rl.bhat

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    Hi Dirac1239, welcome to PF.
    The projectile motion is the combination of vertical and horizontal motion.
    Vertical motion is controlled by g,where as horizontal motion is independent of g. It only depends on total time of flight, and horizontal component of v.
    Now you have collect the kinematic equations, which you have to use here.
    You have to find vertical and horizontal components of velocity of projection.
    Can you do these things?
     
  4. Oct 12, 2009 #3
    I know that, but in order to find total flight time I need to know max height, I order to find max height I need a initial speed which I'm not given.
     
  5. Oct 12, 2009 #4

    rl.bhat

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    You need not know the values. Write the equations in symbols.
    What is the horizontal component of the velocity?
    If x is the horizontal distance and T is the time of flight
    then x = ....?
    What is the vertical component of the velocity?
    What is the time taken by the ball to reach the maximum height?
     
  6. Oct 12, 2009 #5
    Vx=Vi*cos(a) a stands for angle
    okay so total distances in horizontal would be Dx=Vi*cos(a)*t. T is for time, but I have two unknowns how can I solve for any of them.

    Vertical component is Vy=vi*sin(a)
    Max height equals, Viy^2/sg For all of these equations I either need time or initial speed, something I don't have. X for horizontal distance is given, it's 130
     
  7. Oct 12, 2009 #6

    rl.bhat

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    At the maximum height vertical component of the velocity is =...?
     
  8. Oct 12, 2009 #7
    0. But all I still don't know what the max height or the initial speed is. Like if I use Vy^2=Viy^2-2g*Dy, I can't solve it because there is two unknowns.
     
  9. Oct 12, 2009 #8
    How can I solve for something when there is three or even more unknowns.
     
  10. Oct 12, 2009 #9

    Delphi51

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    Hi Dirac. I see the homework helper is offline so I'll offer a thought or two. I'm no P.A.M. Dirac so I have to use the plan for all these projectile problems. I write the horizontal equation: x = v*cos(a)*t
    and two vertical equations: Vy = v*sin(a) - gt and
    y = v*cos(a)*t - .5*g*t^2.
    Then fill in all the knowns and look for a way to make progress. In this case you'll put in the position of the ball as it goes over the fence (130,20). The x= and y= equations above will then constitute a set of two equations with two unknowns (v and t).
     
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