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Speed problem

  1. Sep 17, 2008 #1
    This is my first post here so I want to start off introducing myself. My name is Ian and I'm a working on a degree in athletic training. Ive done pretty well in any physics related classes so far and for the most part I think I'm hanging in there this semester, but I have come across a problem on some homework that has me stumped. Ill be the first to admit that I'm not a huge fan of physics but I am working very hard and paying alot of money for my education and I want to get as much out of that education as I can. I don't want anyone to think I'm just looking for answers to homework, I really want to understand this stuff. So I'm hoping I can make some friends around here who can help me out and I'll do my best to help others as well. Unfortunately the only areas I'll really be helpful in are the areas that I currently teach. Sports and guitar. So if anyone needs to know how to throw a baseball or play a chord, I'd be glad to help. :rofl: Anyways, on to the problem.

    1. The problem statement, all variables and given/known data
    A homerun is hit such a way that the baseball just clears a call 26 m high located 123 m from home plate. The ball is hit at an angle of 38 degrees to the horizontal, and air resistance is negligible. Assume the ball is hit from a height 2 m above the ground. The acceleration of gravity is 9.8 m/s^2

    What is the initial speed of the ball?
    How much time does it take for the ball to reach the wall?
    What is the speed of the ball when it hits the wall?

    2. Relevant equations
    None are specifically given, but I have been trying to use these:
    Displacement of X=Vxt + 1/2at^2 as well as the same for the displacement of Y, with Vy instead of Vx




    3. The attempt at a solution
    I have been working on this for a day or so and haven't seen any success. I started with the equation for the displacement of Y since it is known as 24m. I also arranged time in that equation to become DispX/Vcos(38). After a bit of algebra to get Vo^2 on one side of the equation, I came up with {1/2(9.8ms2(123m)/cos(38)^2)}/{(123m)(tan38)-24m}. I worked that out to be Vo^2=13.48 and Vo = 3.67

    I appreciate any help you guys can give me. Thank you so much for even taking the time to read this over.
     
  2. jcsd
  3. Sep 17, 2008 #2

    LowlyPion

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    Welcome to PF.

    Without working through what you've done, your result doesn't look plausible, so ... let me start to outline an approach.

    You already seem to know that V off the bat has a Vx and a Vy. We will remember that they are Vx=VCosθ and Vy=VSinθ and θ = 38 degrees.

    But for this problem we can break up its flight in several ways and then solve for what we need to know.

    You know that T - Total time - times Vx = 123m
    Time to max height Tup and then Time to top of fence. Tdown. (These will be different times.) But at least we know Ttotal is Tup + Tdown.

    Tup to max = Vy/g (g=9.8m/s2)
    Tdown is found by Hmax - 24m = 1/2*g*Tdown2

    We also know that Hmax = vy2/2*g

    This is a pretty good start and should give you an idea about how to develop a solution.
     
    Last edited: Sep 17, 2008
  4. Sep 17, 2008 #3
    Thank you for getting back to me. I appreciate the welcome. I started over with this is whats happening. Tell me if I am on the right track here please, I feel like I did the same thing again. I took Vx=Vcos38 and Vy=Vsin38 and t= 123/Vcos38 and I put them into the equation of 24=Vsin38*t - 1/2gt^2 and came up with 24 = Vsin38(123/Vcos38) - 1/2g(123/Vcos38)^2. This is where Im getting confused now. I ended up with Vo^2 = {4.9*(123/Cos38)}/123m*tan38-24m
     
  5. Sep 17, 2008 #4
    Alright I solved it, basically I was thinking circles around myself and screwed up some fairly basic algebra. Its funny how something can look so confusing but be so simple. LowlyPion, thank you very much for taking the time to spell that out for me. I don't know how I managed to screw it up initially but I was thinking about everything you said the whole time, something just didn't click until I read over your explanation a few times. Thanks a bunch. I never even thought there may be some kind of forum for help with this kind of stuff but this is an excellent site. I really admire the way you all help each other out and promote learning as opposed to giving away answers. Good stuff. Thanks again. I'm sure Ill be back again
     
  6. Sep 17, 2008 #5

    LowlyPion

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    I got to a different equation than you posted, so I'm not sure how you got there. But yes, care always needs to be taken in making substitutions and moving terms around.

    Whatever the case with your algebra, if it occasioned your learning the way the equations can be interchanged has to be a benefit.

    Good luck.
     
  7. Sep 18, 2008 #6
    The final equation I got when I got it correct was V^2={4.9(123^2/Cos(38)^2)} / {123*Tan38-24} Thats what lead me to the correct answer of 40.7: My mistake was that when I was moving terms around I forgot I needed to square 123 and the Cos value. Is that the way you did it too or is there a more logical way to approach this?
     
  8. Sep 18, 2008 #7

    LowlyPion

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    I don't have those pages in front of me anymore but I recall noting that you didn't have the (123)2 term.

    I hope the extra effort to get it right has paid dividends.

    Good luck.
     
  9. Sep 18, 2008 #8
    Thank you
     
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