Posted by chica on Thursday, May 13, 2010 at 6:40pm.
A diver on a platform 40 feet in height jumps upward with an initial velocity of 5 feet per sec. His height in h feet after t seconds is given by the function h=16t^2+5t+50. What is his maximum height? How long will it take him to reach the surface of the water?

AlgebraQuadratic Applications  Reiny, Thursday, May 13, 2010 at 7:02pm
There is a discrepancy in your problem.
You state the platform is 40 feet high, yet the equation suggests an initial height of 50 feet.
You would find the vertex of the downwards parabola.
An easy way is to find the x of the vertex by b/2a
= 5/(32) = 5/32 seconds
Now sub that back in to find the height when t = 5/32
for the last part,
set 16t^2 + 5t + 50 = 0
and solve for t using the quadratic formula.

AlgebraQuadratic Applications  Damon, Thursday, May 13, 2010 at 7:10pm
I assume you have not covered calculus which would make this easy.
Without calculus you must complete the square to find the vertex of the parabola.
h=16t^2+5t+50
h = 16 t^2  5t  50
50  h = 16 t^2  5 t
50/16 h/16 = t^2 (5/16) t
50/16  h/16 + 25/1024 = t^2 5/16 t + 25/1024
3225/1024  h/16 = (t5/32)^2
we are at the top when t = 5/32
and h = 3225/64 = 50.4
Now what is t when h = 0?
= 16 t^2 +5 t + 50
t = [ 5 +/ sqrt (25 + 3200) ]/32
t = [5 + 56.8 ]/32 = 1.62
ignore that solution, negative time was before we started
t = [5  56.8]/32 = 1.93 seconds
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