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    Hey, I'm stuck with what to do on this q & would appreciate any help/hints:
    4. A girl throws a ball vertically upwards with speed 8 m s^-1 from a window which is 6m above horizontal ground.
    a) Find the greatest height above the ground reached by the ball.
    I did this and got the right answer = 9.27m

    1.5s later she drops a second ball from rest out the same window
    b) find the distance below the window of the pount where the balls meet.
    So for this i got:
    u = 0 m s^-1
    a = 9.8 m s^-2
    t = t - 1.5
    and i worked out that ball 1 gets to the window point on its way down 0.684 seconds before ball 2 is dropped, but not sure on what to do now.
    Thanks
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    1 moment please i must get out ma calculator
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    haha no problem bro
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    okasy first part 9.27 is right ( i know you said so already but i had to be sure)
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    The way to Approach this problem is simple. In the end, what you need is two position equations for each of the ball and you set them equal to eachother. The position equation you need is  y = y_i + V_it+\dfrac{1}{2}at^2

    So the first thing you need to do is find out the (1) Velocity, and (2) the Position of the first ball at 1.5 seconds relative to the ground. Once you find that you plug it into the equation above where y_i = position above ground initially and  V_i is the velocity. a will be -9.8. (don't forget negative sign its important, your initial velocity will probably be nagative you).

    Then you do the same for the second ball .. In this case the equation is simple.  y_i = 6m ; V_i = 0 so it becomes  y = 6 - \dfrac{1}{2}*9.8*t^2 .

    Set them equal to eachother and you get a time when they meet. Then you plug in the time in one of the equations to get the position. .
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    if the original time is t shouldn't the new time for the second ball be t+1.5 insteald of t-1.5?
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    So the position of ball 1 at 1.5 secs is 6.98m above the ground and ball 2 is at 6m
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    the velocity of ball 1 at 1.5 secs is 6.7 m s^-1
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    remember velocity has a direction so because it is falling down, it would be -6.7
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    (Original post by DanielUK_Rocks)
    remember velocity has a direction so because it is falling down, it would be -6.7
    Props to you sir I got the right answer, ill rep you in a mo
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    oh it simple once you see = since the balls collide therefore their displacements is equal - the formula for displacement is s= ut+0.5at^2

    so if s1 is equal to s2 then that means that (assuming s1 is ball 1 and s2 is ball 2) s1= 8(t)+ 0.5(-9.8)(t^2)
    s1=8t-4.9t^2

    ball 2 thrown 1.5s after ball 1 (t+1.5)

    s2= 0(t+1.5) + 0.5(-9.8)(t+1.5)^2 = -4.9(t^2+3t+2.25)

    =-4.9t^2 -14.7t- 11.25

    s1=s2

    so 8t-4.9t^2=-4.9t^2 -14.7t-11.25
    solve for t then substitute for an answer which you would take away from 6

    22.7t=-11.25

    t= -11.25/22.7 =0.495/0.5

    (i think i may have messed up a sign somewhere as t should be positive)
    now substitute into the simplest one 8t-4.9t^2 = 8(0.5)-4.9(0.5^2)

    = 4-1.225 = 2.775 now its 6 metres above the ground but 2.775 metres under the window so therefore 6-2.775 = 3.225m above ground the balls collide!!!! voila!!!! god i hope this is right i could be wrong
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    awww man the other guy beat me to it
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    Yeah thanks for the effort tho & sorry it was wrong lol answer was like 0.1ish. Cheers anyway
 
 
 
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