byerskm2
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?
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anonymous
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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!!
byerskm2
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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
anonymous
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oh you put in the word "dropped" tsk tsk
byerskm2
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I just can't find the corriect value
byerskm2
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sorry that's what the question said
byerskm2
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:P
anonymous
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lets try this
\[a(t)=-32, v(t)=-32t^2+v_0, p(t)=-16t^2+v_0t+224\]
byerskm2
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yes
anonymous
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or more commonly written
\[h(t)=-16t^2+v_0t+224\]
anonymous
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set equal to zero, then put \(t=1\) to solve for \(v_0\)
byerskm2
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208?
anonymous
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\[t^2-\frac{v_0}{16}-14=0\]
anonymous
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oh no your answer will have a \(v_0\) in it
byerskm2
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I plugged in 1 for t and got the 208
anonymous
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oh wait, you are right
put \(t=1\) set equal to 0 solve for \(v_0\)
anonymous
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\[0=-16+v_0+224\]
\[v_0=-208\]
anonymous
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i see the problem
it is negative and you put 208 i bet
byerskm2
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Oh! Darn. I even said it have to be negative in the beginning didn't I? X(
anonymous
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yes you did
byerskm2
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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
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yw
always good to have another pair of eyes