anonymous
  • anonymous
An object is dropped from a tower of height 224 feet, subject to only the constant acceleration of −32 ft/sec2 due to gravity. (d) What should its initial velocity be in order to strike the ground after 1 second?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
clearly the word "dropped" here is inappropriate, since they are asking what hthe initial velocity should be if it is dropped, the initial velocity is 0!!
anonymous
  • anonymous
http://i50.tinypic.com/34ip8a8.png
anonymous
  • anonymous
In order to reach the ground in 1 second the Vi would have to have a negative velocity of some value otherwise it would not be fast enough

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anonymous
  • anonymous
oh you put in the word "dropped" tsk tsk
anonymous
  • anonymous
I just can't find the corriect value
anonymous
  • anonymous
sorry that's what the question said
anonymous
  • anonymous
:P
anonymous
  • anonymous
lets try this \[a(t)=-32, v(t)=-32t^2+v_0, p(t)=-16t^2+v_0t+224\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
or more commonly written \[h(t)=-16t^2+v_0t+224\]
anonymous
  • anonymous
set equal to zero, then put \(t=1\) to solve for \(v_0\)
anonymous
  • anonymous
208?
anonymous
  • anonymous
\[t^2-\frac{v_0}{16}-14=0\]
anonymous
  • anonymous
oh no your answer will have a \(v_0\) in it
anonymous
  • anonymous
I plugged in 1 for t and got the 208
anonymous
  • anonymous
oh wait, you are right put \(t=1\) set equal to 0 solve for \(v_0\)
anonymous
  • anonymous
\[0=-16+v_0+224\] \[v_0=-208\]
anonymous
  • anonymous
i see the problem it is negative and you put 208 i bet
anonymous
  • anonymous
Oh! Darn. I even said it have to be negative in the beginning didn't I? X(
anonymous
  • anonymous
yes you did
anonymous
  • anonymous
I plugged it in :) it's right :P thanks! I knew I had to be right but I'm always missing some stupid detail!
anonymous
  • anonymous
yw always good to have another pair of eyes

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