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Ball throw

  1. Apr 10, 2005 #1
    A ball is thrown with an initial speed of 31 m/s at an angle of 45°.The ball is thrown from a height of 10 m and lands on the ground.

    (a) Find the time of flight.


    I used the following formula:

    Yf = Yo + VoT - (1/2)GT^2

    0 = 0 + (31 m/s sin 45)T - (1/2)(9.8 m/s^2)T^2
    0 = 21.92T - 4.9T^2
    -21.92T = -4.9T^2
    T = 4.47s

    I'm checking my answer for the problem on my school website and it's coming back as wrong. I pretty sure I'm using the correct formula for problem. The math looks right.
  2. jcsd
  3. Apr 10, 2005 #2
    You might want to find the components of the flight in terms of horizontal and vertical flight.

    The Bob (2004 ©)
  4. Apr 10, 2005 #3


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    Of course,that [itex] y_{0}\neq 0 [/itex]...

  5. Apr 10, 2005 #4
    just xaplin why Yf - Y0 is zero?? DId the ball not travel some vertical distance?? I.e. Is the final height same and the initial height??
  6. Apr 10, 2005 #5


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    No,the initial height is 10 m...And he should be careful with those velocity components as well.

  7. Apr 10, 2005 #6
    Well I'm confused with Y (initial) and Y (final). These are the hints that came with the problem:

    We know y initial and y final are both zero since both are on the ground
  8. Apr 10, 2005 #7


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    Have you been reading other problem?

    "(...)The ball is thrown from a height of 10 m and lands on the ground.(...)".

    Does this ring a big bell ?

  9. Apr 10, 2005 #8
    Sorry dextercioby..

    Ok so Y initial = 10m and Y final = 0m

    Yf = Yo + VoT - (1/2)GT^2
    0m = 10m + (31 m/s sin 45)T - (1/2)(9.8 m/s^2)T^2
    T = 6.5s

    Tycho is not taking my answer.
  10. Apr 10, 2005 #9
    I got 4.9 seconds for it to hit the ground.

    The Bob (2004 ©)
  11. Apr 10, 2005 #10
    Unfortunate, start by finding the component of the vertical. This is all you really need to solve the question.

    The Bob (2004 ©)
  12. Apr 10, 2005 #11
    Have I gone mad? Did unfortunate just post or am I dreaming???

    The Bob (2004 ©)
  13. Apr 10, 2005 #12
    Our posts crossed. I asked what steps you took to get 4.9s. I deleted it thinking that you were going to explain the steps.
  14. Apr 10, 2005 #13
    Fair enough well the explaination will be in the next post.

    The Bob (2004 ©)
  15. Apr 10, 2005 #14
    The vertical component of the ball can be expressed in the form:

    [tex]s = ut + \frac{1}{2}at^2[/tex]

    u = sin 45° x 31 = 22 ms-1, a = -9.8 ms-2 which will leave an expression for t relative to the ball's position (s).

    This gives s = 22t - 4.9t² + 10 (the 10 has come from the fact that the ball has started 10 metres above the ground).

    Now when s = 0 the ball has landed (yes?). This means we have a quadratic equation : 0 = 22t - 4.9t² + 10. Rearrange this so that the squared function is positive and you have 4.9t² - 22t - 10 = 0.

    Then it is a case of using the quadratic equation:

    [tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

    This (for t instead of x) will leave you with either 4.9s or -0.41s. The negative answer cannot be what you are looking for (as this answer will give you how long ago the ball would have needed to be throw from the ground to get to 10 metres above the ground).

    Hence, the answer is 4.9 seconds.

    Hope that helps. :smile:

    The Bob (2004 ©)

    P.S. You had the idea but didn't seem to know where to take it. :smile:
    Last edited: Apr 10, 2005
  16. Apr 10, 2005 #15
    Genius.. I'm soaking it all in. :) Now to attack the rest of the ball throw problem.
  17. Apr 10, 2005 #16
    Good Luck then :wink:

    The Bob (2004 ©)
  18. Apr 10, 2005 #17
    Ok I'm having a problem finding the speed at impact. I know it's the magnitude of Vx and Vy.

    I have Vx as Vi cos 45 = 21.92

    I'm having trouble finding Vy.
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