SOLUTION: The height of a projectile, over time, launched from an initial height of 50 feet can be modeled by the quadratic function, h(x) = -16t2 + 96t + 50, where h(x) is the height of the

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The height of a projectile, over time, launched from an initial height of 50 feet can be modeled by the quadratic function, h(x) = -16t2 + 96t + 50, where h(x) is the height of the      Log On


   



Question 725361: The height of a projectile, over time, launched from an initial height of 50 feet can be modeled by the quadratic function, h(x) = -16t2 + 96t + 50, where h(x) is the height of the projectile and t is time, in seconds. If the maximum height reached by the projectile is 344 feet, how long does it take for the projectile to reach its maximum height?
Thanks!

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The height of a projectile, over time, launched from an initial height of 50 feet can be modeled by the quadratic function, h(x) = -16t2 + 96t + 50, where h(x) is the height of the projectile and t is time, in seconds. If the maximum height reached by the projectile is 344 feet, how long does it take for the projectile to reach its maximum height?
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max height is the vertex of the parabola, t = -b/2a
Why do you think it's 344 feet?
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PS Use ^ (Shift 6) for exponents.
h(x) = -16t^2 + 96t + 50
PPS Should be h(t)