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Physics: Projectile Motion with different starting and ending heights?

  1. Jan 31, 2009 #1
    All the physics problems I have worked with thus far assume stating and ending at the same height. Any help is how to solve this is appreciated :-)

    A baseball player throws a ball from height 1.5 meters at an angle of 30 degrees and an initial velocity of 20 m/s. He is trying to hit a pole 20 meters away which is 1.7 m tall. Forget about air drag, etc

    *how long does it take the ball to travel 20 meters (in the horizontal)
    *what is the height of the ball when it reaches the pole?
    *what is the final velocity of the ball?
    *what is the final angle (assuming the ground is perfectly flat of course)

    To find the time does this sound right?: t = (20meters)/(20m/s * cos (30))

    This is where I am stuck... I would think height at 20 m is just the y component analyzed at 20 but I'm not sure about that. The final velocity I have not idea on and the final angle would probably be the sum of the x and y components although I don't know how to manipulate the formuli to get that :-( Any help is greatly appreciated :-)
    Last edited: Jan 31, 2009
  2. jcsd
  3. Jan 31, 2009 #2


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    From any source can you find the equation of projectile motion which gives you the relation between x, y and angle of projection theta?
  4. Jan 31, 2009 #3
    sure, do you mean y = (tan angle)x - ((gx^2)/(2(v(initial)cos(angle)^2) ?

    x would be the distance to my object (in this case 20m)
    I know the angle
    I know acceleration due to gravity
    The velocity I presume is the initial velocity before you break it into its components?
  5. Jan 31, 2009 #4
    ok, find time taken to travel the 20m horizontaly using trig to find horizontal velocity
    speed = distance/time ==> time = distance/speed
    i would take 1.5m as being height=0 so the pole top is 0.2m then just add your 1.5 back in at the end, so esentualy your working out the change in height :]
    now use trig again to find verticle velocity, you now know time to reach the pole.
    use x=ut+(1/2)a(t^2) remember to keep track of your + and - signs
    x=dissplacement, u=initial velocity(in this case vertical), a=acceleration, t=time

    final velocity, horizontal doesnt change due to no horizontal forces hence obeys newtons 1st law.
    verticaly however use V^2=u^2+2ax
    v=final velocity.
    use c^2=a^2+b^2 (pythagoras) to combine ur horizontal and verticle components of the velocity.
    also use trig with these velocity componetns to find the angle.
    this is all true if the ball does reach the stick if the ball hits the floor befor it reaches the stick then your value for t will be incorect, if after working out your t like i stated you also use
    v^2+u^2+2ax in the verticle to find the final verticle velocity (remember x is dissplacement not distance so here x would equal 1.5m) then use
    v=u+at to find time. if this time is less than the time you calculated horizontaly then the ball wont reach the pole and use v=x/t ==> x=vt horizontal to see how far the ball travels

    Hope i helped.
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