a water balloon is tossed into the air with an upward velocity of 25 ft/s its height h(t) in ft after t seconds is given by the function h(t) = -16t^2+25t+3 after how many seconds will the balloons hit the ground?

ANSWERS

2015-11-07 18:56:42

when the height is 0 that is when it hits the ground solve 0=-16t²+25x+3 cant factor so use quadratic formula for 0=ax²+bx+c [latex]x=frac{-b+/-sqrt{b^2-4ac}}{2a}[/latex] so for 0=-16t²+25t+3 a=-16 b=25 c=3 so [latex]x=frac{-25+/-sqrt{25^2-4(-16)(3)}}{2(-16)}[/latex] [latex]x=frac{-25+/-frac{625+192}}{-32}[/latex] [latex]x=frac{-25+/-frac{817}}{-32}[/latex] it has to be a positive time so it will hit the ground after [latex]frac{25+sqrt{817}}{32}[/latex] seconds

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