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Ball is thrown vertically upward from a window

  1. Apr 25, 2016 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown 2.8m/s vertically upward from a window that is 3.6m above the ground. calculate the amount of time the ball spends in the air.

    2. Relevant equations
    quadratic equation
    d = Vi(t) + 1/2(a)(t)2

    3. The attempt at a solution
    d = Vi(t) + 1/2(a)(t)2
    3.6m =2.8 m/s[up](t) + 1/2(-9.8m/s2[up])(t)^2
    0 = -4.9m/s2[up](t)2 + 2.8m/s[up](t) - 3.6m
    __________________
    into the quadratic formula:

    t = [-(2.8) +- sqrt(2.82 - 4(-4.9)(-3.6))] / 2(-4.9)

    i end up with a negative under the root sign?
     
  2. jcsd
  3. Apr 25, 2016 #2
    The d in the formulae Is the net displacement . Since d initial is 3.6 and d final is 0, d net becomes -3.6
     
  4. Apr 25, 2016 #3
    wait why is the final 0? isnt the displacement of the person always 3.6 [up] ?, also to find total displacement it is final minus initial right?
     
  5. Apr 25, 2016 #4

    SteamKing

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    You're getting ahead of yourself here.

    The ball starts out at 3.6 m above the ground because that's the point from which it is thrown upward. It travels an as yet unknown distance above that starting point.

    What you need to do is to calculate how long it takes for the ball to stop rising after it is thrown upward. Then you can find the maximum height the ball travels above the ground. After the ball stops rising, then it must free fall back to the ground.

    The problem is looking for the total time the ball spends aloft, going up and coming back down.

    This is more than a simple plug and chug problem. You should make a simple sketch so that the different parts of the ball's flight are clear.
     
  6. Apr 25, 2016 #5
    I am confused, soo i drew it out and i have 3 points, the starting point (1) then the maximum point (2) and then the ending point (3), but i don't know what to do from here. all i think i know is that it starts accelerating negatively between 1 and 2 and then positively between 2 and 3?
     
  7. Apr 25, 2016 #6
    U must decide what is positive and what's negative. Since u have taken acc due to gravity as negative, going up will be considered as positive displacement . The initial position is 3.6 m ABOVE the ground. And the final position of the ball is on the ground. So the initial d becomes +3.6. And the final d becomes 0. What's the net displacement?final - initial. Hence the d to be used in the equation is -3.6.
     
  8. Apr 25, 2016 #7
    ahh i understand the 3.6 now, so when i put it in the formula i got approx. 1.2s. however am i doing this wrong because SteamKing above^ said i cant just put it in the formula like that, so i am confused as to what to do, i tried looking in the textbook for an example like this but cannot find one.
     
  9. Apr 25, 2016 #8

    SteamKing

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    Take the first part of this problem.

    You stick your hand out the window and throw the ball up. How long does it take for the ball to stop rising, if you throw it up at an initial velocity of 2.8 m/s?

    For now, forget about the window being 3.6 m above the ground and concentrate on how long it takes the ball to stop rising.
     
  10. Apr 26, 2016 #9
    ok so i got .28s for the first part by using the (vf-vi)/a
     
  11. Apr 26, 2016 #10

    haruspex

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    I disagree completely. The acceleration is constant throughout. Finding the time to reach the apex is unnecessary extra work. It is simple plug and chug.
    @totomyl , your 1.2s in post #7 is correct, and obtained validly.
     
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