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General Physics 1

  1. Jan 23, 2005 #1
    A rock is thrown vertically upward from ground level at time t=0. At time t=1.5s it passes the top of a tall tower, and 1.0s later t reaches its maximum height. What is the height of the tower? ....... is this the equation i would use....... delta(Y)= Vo(t) + 1/2(a)(t2)...........with Vo= 0 , t=1.5, a= 9.8m/s2.....with delta(Y) in Height......= 11.025meters......need help on this one :confused:
     
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  3. Jan 23, 2005 #2

    dextercioby

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    Use this equation
    [tex]v_{fin}=v_{0}-gt [/tex]

    to find the initial velocity (which is not zero),knowing the total time of flight...
    Then this
    [tex] h=v_{0}t-\frac{1}{2}gt^{2} [/tex]

    to find the height...

    Daniel.
     
  4. Jan 23, 2005 #3
    hey thanks so much for helping me out.....for the answer i got h= 30.625m......iam not sure if thats the answer you would have got.....but this is my setup..... Vo= V + gt..... i got 24.5m/s2......then i used Vo-1/2gt2......which i got it to be = 30.625m.....

    Thank you for your help.....

    Krishna Patel
     
  5. Jan 23, 2005 #4

    dextercioby

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    Do u agree that
    [tex] h_{tower}=24.5\frac{m}{s}\cdot 1.5s-\frac{1}{2}9.8\frac{m}{s^{2}}\cdot (1.5s)^{2} [/tex]

    ??

    Do you understand what i did??

    Daniel.
     
  6. Jan 26, 2005 #5
    yeah i see what i did wrong i plugged in 2.5s....instead of 1.5 which it took the time to pass the tower.....i thought it would be the where it reaches its max height for time......but i see how its done...thanks so much....
     
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