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    Can people help? I will Rep one of the responders when i next log on!

    1) A stone is catuplulted vertically upwards with a velocity of 25ms^-1 from a point 2m above the ground . Find a) the velocity when it hits the ground and b) the time it takes to reach the ground

    2) Water from a fountan rises to a height of 6m by modelling the drops of water s particles . Find the speed of the water as it leaves the nozzles

    Thanks all of you
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    1)a. u=25 v=? s=-2 a=-9.81
     v^2=u^2+2as = 625+(2\cdot -2\cdot -9.81 \cdot) = 664 \\ v= 25.8

    b. s=-2 u=25 a=-9.81
     s=ut+0.5at^2 = 25t-4.9t^2
     4.9t^2-25t-2=0

     t=5.2
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    b)

    s = 6
    u =
    v = 0
    a = -9.8
    t =

    v^2 = u^2 + 2as
    0 = u^2 - 2*6*9.8
    u^2 = 12*9.8
    u = root(12*9.8)
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    2. V^2= U^2 + 2AS V=0 as its maximum height so U^2 = 2AS as A will be - in this case. U = root ( 2 x 9.8 x 6) U = 10.84 to 2d.p. i think thats right.
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    (Original post by jameso88)
    2. V^2= U^2 + 2AS V=0 as its maximum height so U^2 = 2AS as A will be - in this case. U = root ( 2 x 9.8 x 6) U = 10.84 to 2d.p. i think thats right.
    Seconded
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    (Original post by jameso88)
    2. V^2= U^2 + 2AS V=0 as its maximum height so U^2 = 2AS as A will be - in this case. U = root ( 2 x 9.8 x 6) U = 10.84 to 2d.p. i think thats right.
    The other way is to think in terms of conservation of energy:
     \frac{1}{2}mv^2=mgh , which will give the same result.
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    Each to their own.

    OP you should really be doing these yourself otherwise you will never learn it.
 
 
 
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