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April 1, 2015

April 1, 2015

Posted by **chica** on Thursday, May 13, 2010 at 6:40pm.

- Algebra-Quadratic Applications -
**Reiny**, Thursday, May 13, 2010 at 7:02pmThere 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.

- Algebra-Quadratic Applications -
**Damon**, Thursday, May 13, 2010 at 7:10pmI 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 = (t-5/32)^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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