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Can someone help me

  1. Aug 31, 2005 #1
    A foul ball is hit straight up into the air with a speed of about 22 m/s.
    (a) How high does it go?
    ? m
    (b) How long is it in the air?
    ? s

    How do I solve this
     
  2. jcsd
  3. Aug 31, 2005 #2

    HallsofIvy

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    You use basic formulas that you surely already know. The acceleration due to gravity (near the surface of the earth) is -9.1 m/s2.

    v(t)= v0- 9.1 t
    where v(t) is the velocity of the ball at time t seconds after the ball was hit staight up and v0 is the initial velocity (here 22 m/s).

    h(t)= h0+ v0t- 4.95 t2
    where h(t) is height of the ball at time t seconds after the ball is hit and h0 is the initial height of the ball. Here you might take it to be 0 although, really, it should be about the shoulder height of the batter- which is not given.

    As long as the velocity is positive, the ball is still going up- not yet at its highest point.
    When the velocity is negative, the ball is coming back down- already past its highest point.
    What do you think the velocity is at the highest point?
    Put that into the equation and solve for t to find when the ball is at its highest point. Now use that time in the height equation to find that height.

    "How long is it in the air?"

    It's in the air until it gets back to the ground! What is its height when it hits the ground? If you put that into the height equation and solve for t what do you get?
     
  4. Aug 31, 2005 #3

    TD

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    Just a small note, HallsofIvy probably meant g = 9.8 m/s² (the 1 probably came from 9.81).
     
  5. Aug 31, 2005 #4
    I got the first part. But I am still confused on the second part. I cant find the velocity of the ball at its hightest point.
     
  6. Aug 31, 2005 #5
    Nevermind the answer is 4.5 seconds right?
     
  7. Aug 31, 2005 #6

    mezarashi

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    Lay out the BIG FOUR (kinematics equations). There are only 5 variables: acceleration, time, initial velocity, final velocity, and distance. You will have to be given atleast 3 of these 5 variables. Pick out the equation in which there is only 1 unknown and solve.

    How high does it go?
    Vinitial = 22m/s, a= gravity = -9.8m/s/s, Vfinal = 0 m/s (at its peak)
    d = ?

    How long does it take to get to the ground? Using the conservation of energy the velocity when it reaches the ground again must be equal to when it went up, Vfinal = -22m/s, a = -9.8m/s/s, Vinitial=22m/s, t=?
     
  8. Aug 31, 2005 #7

    HallsofIvy

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    The answer to what question? :smile:

    As far as "I cant find the velocity of the ball at its hightest point." I was trying to give you that: as long as the velocity is positive, the ball is still going up and so is not at its highest point! If the velocity is negative, the ball is now going down and so is past its highest point! The velocity at the highest point can't be positive or negative so it must be 0.

    Yes, I meant to write that g= -9.81. Don't know where the "8" got to!

    If v(t)= 22- 9.81t= 0, then 9.81t= 22 so t= 22/9.81= 2.24 s when the ball is at its highest point.
    Now use h(t)= 22t- 4.91t2 to find the actual height at that time.

    Of course the ball ends its flight when h(t)= 0 again. You can solve
    22t- 4.91t2= 0 to find the time that happens.
    22t- 4.91t2= t(22- 4.91t)= 0 so t= 0 or t= 22/4.91= 4.48 seconds (4.5 to two sig. figures). Of course, because of the symmetry, since we already knew that the ball took 2.24 s to reach its highest point, we could have argued that it will take the same time coming back down: total flight 2.24+ 2.24= 4.48 s.
     
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