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Homework Help: Trouble with ball being thrown up in the air problem

  1. Feb 6, 2016 #1
    1. The problem statement, all variables and given/known data
    The problem reads: A person throws a ball up in the air with an initial velocity of 15m/2. Calculate the max height reached and how long it is in the air for.

    2. Relevant equations
    x = x0 + v0t + 1/2at^2

    3. The attempt at a solution

    I need to calculate the how long the ball is in the air for

    My book gives me the answer but only does it in one go

    I'd like to divide it into two parts, when it reaches max height, and when it comes down

    I keep doing the calculations and I keep getting it wrong

    My book does uses y= y0+v0t+ (1/2)*(at^2) to calculate the entire time the ball is in the hair

    I would think to use that formula to get the time of the ball until it reaches the max height, like this


    The problem when I do that is that I get a negative value inside of the square root

    So I'm stuck there

    Then to calculate the time between max height and returning to the thrower's hand, I'd do 0=11.5+15t+4.90t^2 (9.80 is positive here because the object is going in the downward direction, right??)

    so I would like if someone could tell me what I'm doing wrong

    Other people use the other formulas in the book but I don't understand why and I wouldn't think to use them, I would think to use the one I used in the description because when I check what data I know, it lines up with that formula, so please try to use that formula in order to explain because I don't see the logistics in using the other formulas

    Thank you
  2. jcsd
  3. Feb 6, 2016 #2


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    This should be okay. You shouldn't get a negative inside the square root.

    Take a look at that equation you've just written: 11.5 is positive, 15 is positive (equates to the ball moving upward at 15m/s at that point) and gravity is positive (which means it accelerates the ball upward). That ball is just going to go up and up!

    You don't have to split the motion into two halves, but it's not a bad exercise to do this, so it's worth getting it right.
  4. Feb 6, 2016 #3
    Yes you can definitely break the motion into two parts.
    Part 1) When the ball is going up. Use the first equation of motion to find time to reach max height. Use third equation of motion for finding the max height.
    Part 2) When ball is falling down. Use the same procedure
    Add the times acquired through both parts and get time of flight.
    Just take care of the signn conventions.
  5. Feb 6, 2016 #4
    But when I do the quadratic do get a negative value inside of the square root because its 15- (4*(-4.90)(-11.5))

    And yes I made a mistake, it should be 0=11.5-15t+4.90t^2
  6. Feb 6, 2016 #5
    That's what my book also did.
    I don't understand why I would use the first equation of motion because it leaves out the fact that the ball traveled up to 11.5 meters, unlike the second equation of motion.
  7. Feb 6, 2016 #6


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    You're just putting numbers into equations any old way, without thinking about what you are doing. That's a ball starting 11.5m off the ground, travelling at -15 m/s (i.e. downward) and being accelerated upwards at 9.8 m/s^2 by gravity. You need to think about what these equations actually mean.

    You also need to check the quadratic formula.
  8. Feb 6, 2016 #7
    The ball is thrown with an initial velocity of 15m/s and reaches a maximum height of 11.5. Why would the velocity be negative if its being thrown upward?

  9. Feb 6, 2016 #8


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    You've got it negative in your equation. Moreover, you've put ##v = -15 m/s## at a point where ##v = 0##. And you've got gravity as an upwards force in your equation.

    And, yes I can see your attempt at the quadratic formula is wrong. I meant for you to look up the correct formula!
  10. Feb 6, 2016 #9

    I have -15 because if I just plug it into the equation of motion formula it's 11.5= 0 +15t -4.90t^2

    When I put everything on one side and equal it to cero the signs change.

    And yes I did forget to square the b, but I didn't in my previous attempts, and I still get a negative argument
  11. Feb 6, 2016 #10
    My main issues is that when calculating the time of flight of the ball until it reaches the maximun height, people seem to always use the v=v0+at formula, and I don't understand why, why would they if they are not using the height, and if that is the formula to use to get the times of flight of each "section" so to speak, how can I use that exact same formula to calculate the time of flight after it reaches the maximum height when the initial velocity is zero since it was at the top?! I just don't see why that formula is being used when the max height is being left out and how it can be used for the time of flight after the max height.
  12. Feb 6, 2016 #11


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    I can see you've got ##15^2## in the equation now, which is correct. But, it's still negative! I better tell you the answer. The disciminant should be 0, but because of a rounding error it's coming out very small and negative. That was tricky!
  13. Feb 6, 2016 #12


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    1) You don't need to calculate the height to calculate the time of flight. The time of upward flight is determined by the initial velocity and the deceleration due to gravity. In general, it's often better to get a time than a distance. With experience, you may find time is a more useful thing to know.

    That's what I would have done: ##t_{up} = \frac{15}{9.8} s##

    2) If you want the time up and down. The quickest way is to note that by symmetry of the motion ##t_{up} = t_{down}##.

    Another way is to look for the time when ##y = 0##:

    ##0= 15t - 4.9t^2 = t(15-4.9t)## so we need ##t(15- 4.9t) = 0##

    One solution is ##t=0## (that makes sense) and the other solution is ##t = (15/4.9) s##, which is simple, avoids quadratic equations and confirms that the total time is twice the time to reach the maximum height. I.e. ##t_{total} = 2 t_{up}##

    And everything ties up nicely.
  14. Feb 6, 2016 #13
    Ok I understand now, thank you so much!
  15. Feb 6, 2016 #14
    If you want to solve it by your way then you should do the following
    Get the maximum height first
    By 15/(2*9.8)
    Which gives you approximately 11.479 m
    Now place it in the equation: 11.479 = 15*t - 0.5 * 9.8* t2
    Make it look better :P: 4.9* t2 - 15*t +11.4795 = 0
    In this way of solving you have to put the values exactly as it is. Any slight difference it might change the value of time.
    You will get the time of 1.53 s (One solution only) when you solve it.
    If you want to double it to get the flight time it is up to you if not
    Then you have another equation to handle:
    It is at 11.479m right now so put that in the equation and as it is falling without any initial speed
    0 = 11.479 - 0.5 *9.8* t2
    Solve it and you will find the time of 1.53 s.
    Sum it and bravo you got it :D
    As Perok said, That way he used is much much simpler than what you are doing. There are a lot of ways to use you dont have to follow anything just do what you want.

    Another way is, Use this equation Vf = vi + at
    when it is back from the maximum height to the position of throwing its velocity is -15
    -15 = 15 - 9.8*t
    Gives you 3.06 s (Time of the whole flight)
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