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Help with freefall problem please.

  1. Jan 17, 2006 #1
    A fish is able to jump vertically out of the water with a speed of 4.5 m/s. How much time does it take for the fish to pass a point 0.4 m above the water on the way down?

    Can someone please post how to do this? I know you need to solve for t by using the quad. equation but I have done it many times and cannot get the correct answer. Can someone please post a step by step solution so I can see where I am going wrong? Thanks in advance.
     
  2. jcsd
  3. Jan 17, 2006 #2
    We don't do that here.

    However, if you show the work you've already done on this problem, people will point out your mistakes and help you solve your own problem.
     
  4. Jan 17, 2006 #3
    Okay

    the equation i used was y-yo=vot+(1/2)at^2.

    yo is zero because the fish starts at the water.
    y is the distance that the problem gives you which is .4m.
    a is 9.81, which is gravity.
    vo is 4.5m/s.

    But I am stuck. Do you use -9.81 since the fish is falling or do you first have to solve for the time it takes to go all the way up to zero velocity and then solve again for time to get to .4m above the water and add that number to the previous up to zero velocity?
     
  5. Jan 17, 2006 #4
    Yes, the acceleration due to gravity will be negative if you use up as a positive reference. So, you'll get a quadratic with two answers, and pick the larger one.
     
  6. Jan 17, 2006 #5
    You need to know that the direction of the acceleration is not positive, but negative if your velocity is positive.

    Also, you might want remember that your equation should yield two answers and you have to know which one to use for the fish 'falling' past 0.4m
     
  7. Jan 17, 2006 #6
    I think you've got the right idea.

    You want to split it into two problems; one for going up, and the other for coming back down.
     
  8. Jan 17, 2006 #7
    No, no. You only need to use the function for displacement.
     
  9. Jan 17, 2006 #8
    Using the formula I posted above I still get the wrong answer. I just plug the numbers in and solve for t right? I'm a little confused as to whether I need to approach the problem differently. Do I need to do 2 separate calculations, one for the the fish going up and then one for it coming down?
     
  10. Jan 17, 2006 #9
    Did you get .819s?

    Show us what you did.
     
  11. Jan 17, 2006 #10
    Yes that is what I got but it's wrong. So I'm doing something wrong here.
     
  12. Jan 17, 2006 #11
    Seems like I would need more info to solve this problem than what I have been given in the problem. Too many variables?
     
  13. Jan 17, 2006 #12
    you have enough information to use the equations
     
  14. Jan 17, 2006 #13
    Well what am I doing wrong? Is it a 2 part problem?
     
  15. Jan 17, 2006 #14
    I'll give you a dollar if I can have your brain.
     
  16. Jan 17, 2006 #15
    a=4.905 or -4.905?
    b= 4.5
    c= .4
    plug those into quadratic eq.?
     
  17. Jan 17, 2006 #16
    Huh? What's their answer?
     
  18. Jan 17, 2006 #17
    Nevermind. I figured it out. Use velocity formula to find time at zero v. Use that time to solve for distance traveled at 4.5m/s over the t of .459. Then do the freefall formula to solve for time at .4m then add that time to previous t solved for. It worked. Makes sense intellectually, just needed to figure out the logistics of how to approach it. Thanks.
     
    Last edited: Jan 17, 2006
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