Question 285385
Since the height of a dive is measured from the water, her height when she hits the water will be zero. So the equation at the moment she hits the water will be
{{{0=-4.9t^2+8.8t+3}}}
Since this does not factor we'll use the Quadratic Formula with "a" = -4.9, "b" = 8.8 and "c" = 3:
{{{t = (-(8.8) +- sqrt((8.8)^2 - 4(-4.9)(3)))/2(-4.9)}}}
Simplifying we get:
{{{t = (-8.8 +- sqrt(77.44 - 4(-4.9)(3)))/2(-4.9)}}}
{{{t = (-8.8 +- sqrt(77.44 + 58.8))/(-9.8)}}}
{{{t = (-8.8 +- sqrt(136.24))/(-9.8)}}}
{{{t = (-8.8 +- 11.6721891691318985)/(-9.8)}}}
Now we should split this into the two expressions for t:
{{{t = (-8.8 + 11.6721891691318985)/(-9.8)}}} or {{{t = (-8.8 - 11.6721891691318985)/(-9.8)}}}
{{{t = (2.8721891691318985)/(-9.8)}}} or {{{t = (-20.4721891691318985)/(-9.8)}}}
Since the first answer is negative we will reject it. (It makes no sense to deal with any time <i>before</i> she started her dive!) Dividing out the second solution we get:
t = 2.0889988948093774
So she hits the water approximately 2 seconds after starting her dive.