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Homework Help: 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

    3. The attempt at a solution

    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.


    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


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    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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