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Homework Help: Quadratic equation ball 72km/h

  1. Jun 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown at 72km/h speed from top of a building. Building is 125m tall.

    2. Relevant equations

    Distance travelled before it hit the ground is as follows:

    s = Ut + 0.5 gt^2

    3. The attempt at a solution

    Using quadratic equation:
    x = -b±b^2 - 4ac / 2a

    The question is for part one:
    Find the time for the ball to drop to fifth of the height of the building.

    Now I'm not sure if that means to 25m or 100m?!
    For 25m = 1 second
    For 100m = 3.3852

    I got these from the quadratic equation. Do they seem right?

  2. jcsd
  3. Jun 30, 2013 #2
    The question is not clear. Whether the ball is thrown up or thrown down?

    This is wrong. There is square root missing.
  4. Jun 30, 2013 #3
    Sorry again I clicked the root sign but it couldn't of worked. Here is the equation again:

    x = -b ± √b^2 - 4ac / 2a

    The question is unclear on the "fifth" part. Do they mean from top or bottom?

    Anyway I calculated it from all heights:

    At 25m = 1 second
    At 100m = 2.899 seconds
    At 125m = 3.3852 seconds

    Seems right?
  5. Jun 30, 2013 #4
    The question is clear here. They asking from the top that is 25m.

    Should i solve it and check it? I am not going to do that. It would better if you tell me what you put the values of U, g and S in this equation. I think this would be enough to enough to tell whether you did it correctly or wrong.

    s = Ut + 0.5 gt^2
  6. Jun 30, 2013 #5
    U = 20ms (metres per second) (initial velocity of 72 Km/h I converted to ms)
    g = 10ms (this is given in the question (g = acceleration due to gravity 10ms). Without which I'd of put 9.81ms, but I assume they allowed for ball mass?)
    s = distance, so 25 metres, 100 metres and 125 metres (although by the sounds of it I can drop the 100 metre one)
  7. Jun 30, 2013 #6
    Ah, of course! I got so wrapped up in the quadratics and factoring today I forgot I had the original equation to "plug" my t into and check the solution I got was correct :-)
  8. Jun 30, 2013 #7
    Everything looks so damn correct. Good work!

    You can always press the thanks button if you find my help useful :)
  9. Jun 30, 2013 #8
    Done the thanks button. Brilliant mate THANK YOU!!!!!!!!!!!!!!!!!!!!!!
  10. Jun 30, 2013 #9
    As the question doesn't state whether the ball is thrown upwards or downwards. You have taken the downward case.

    Question: How will the values change if the ball is thrown upwards?
  11. Jun 30, 2013 #10
    The question does say downwards, sorry for my poor copying!
    I'm done for the day, but if it had been upwards g would of worked "against", not "for" the acceleration.
  12. Jul 1, 2013 #11
  13. Jul 1, 2013 #12
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