<PRE><B>Question 763472</B>
The quadratic equation to use is
s(t) = -4.9*t^2 +v(0)*t +h(o)
note that we use 4.9 since we are working in meters, also h(o) = 8 meters
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now we need to calculate v(o) which is the launch velocity and
we use rate * time = distance to calculate the rate which is v(0), so
v(0) * .30 seconds = .4 meters and 
v(0) = .4 / .30 = 1.33 meters/sec
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now returning to the quadratic above we have
0 = -4.9*t^2 +1.33*t + 8
solve for t using the quadratic formula
t = (-b+sqrt(b^2-4ac)) / 2a = (-1.33 +sqrt(1.33^2 -4*(-4.9)*8)) / 2*-4.9 = -1.15
t = (-b-sqrt(b^2-4ac)) / 2a = (-1.33 +sqrt(1.33^2 -4*(-4.9)*8)) / 2*-4.9 =  1.42

therefore she reaches the water in 1.42 seconds
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