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Projectile Motion Softball Question

  1. Dec 17, 2007 #1
    1. A softball is tossed into the air at an angle of 51.3 degrees with the vertical (that would be 38.7 degrees with the horizontal). The initial velocity is 17.3 m/s. What is the maximum height of the softball?



    2. Relevant equations
    Vx=VcosTHETA
    Vy=VsinTHETA
    t=v/a
    d=1/2(a)(t^2)
    v=d/t


    3. The attempt at a solution
    17.3sin51.3=13.50
    t=13.50/9.8=1.378
    d=(1/2)(9.8)(1.378^2)=9.3


    I've tried this problem 9 different ways (the one above is most recent) and I can't seem to figure it out. I have one more try to submit before I cannot answer the question anymore. My other answers include: 10.817, 13.5, 27.17, 27.67, 22.17, 573.305, 230.496, and 5.96944. If anyone could help me with this question I would really appreciate it! Thank you for your efforts!! :)
     
  2. jcsd
  3. Dec 17, 2007 #2
    the correct answer is last one in your list of answers 5.96 m. observe the way you did that.

    17.3sin51.3=13.50

    this here is wrong, you need the sin of 38.7. the angle to the horizontal is what you need.
     
    Last edited: Dec 17, 2007
  4. Dec 17, 2007 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You want the vertical component of velocity for your v if you are going to use t=v/a. That's 17.3m/sec*cos(51.3) or 17.3m/sec*sin(38.7). Your effort of 5.96944m looks very correct. You may have a significant figures problem if it's not acceptable.
     
  5. Dec 17, 2007 #4
    I'm using the program Web Assign and for this particular problem, the amount of sig figs is not significant... I redid the problem and got 5.969436459 as the exact answer. Did I round incorrectly when I submitted 5.96944? I'm so confused. Thank you for responding! :)
     
  6. Dec 17, 2007 #5
    you want to three sig digits, that is the lowest ammont given in the question.
     
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