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h(t)=-16t^2+voT+ho. Initial velocity is 64ft/s. The rocket hits the ground in 4 seconds. If the rocket reached its maximum height (vertex) t=2 seconds, what is the value of maximum height?
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I'm pretty sure the initial height is zero, so you don't even count it
4 = -16t^2+64t Now we just solve for t
-16t^2+64t - 4 = 0
-32t^2-256t-16
never mind..im suppose to factor right?
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-4(4t^2-16t+1) = 0 4t^2-16+1 = 0
is it 96 feet?
4t^2+(-8 + 2sqrt{15})x + (-8 - 2sqrt{15})x + 1 = 0
confused...
ok
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Belay that
Can you take a look at this Hero when you get a chance. Thanks http://openstudy.com/?F47433677217634HJ3W=_#/updates/4eb9de7ce4b0fc85a8ca99ac
belay?
okay I got it..thanks
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tht was a complicated problem..thanx for helping
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