Question 76800
A gun fires a bullet almost straight up from the edge of a 150-foot cliff. If the bullet leaves the gun with a speed of 394 feet per second, its height at time t is given by h(t)=-16t^2+394t+150, measured from the ground below the cliff. When will the bullet land on the ground below the cliff? (Hint: What is its height when it lands?
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It's height will be 0 when it lands, right?  Therefore h = 0, so we have:
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-16t^2 + 394t + 150 = 0; solve for t, the positive solution is what we want
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Use the quadratic formula, a=-16; b=394; c=150
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
:
{{{t = (-394 +- sqrt( 394^2- 4 * -16 * 150 ))/(2*-16) }}}
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{{{t = (-394 +- sqrt(155236 - (-9600) ))/(-32) }}}
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{{{t = (-394 +- sqrt(155236 + 9600 ))/(-32) }}}
:
{{{t = (-394 +- sqrt(164836 ))/(-32) }}}
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1st solution:
{{{t = (-394 - 406 )/(-32) }}}
:
{{{t = (-800)/(-32) }}}
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t = +25 seconds, this is only solution we care about. 
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Make sense to you? I hope it did.