A ball is thrown across a field from a height of 6 feet above the ground. The position y of the ball above the ground is modeled by y=(-32)/(400)x^2+x+6, where x is the distance (ft) that the ball has traveled horizontally. At what exact location x does the ball reach its maximum height?

unjulpild9b

unjulpild9b

Answered question

2022-09-23

A ball is thrown across a field from a height of 6 feet above the ground. The position y of the ball above the ground is modeled by y ( x ) = 32 400 x 2 + x + 6, where x is the distance (ft) that the ball has traveled horizontally. At what exact location x does the ball reach its maximum height?

Answer & Explanation

Kaiden Stevens

Kaiden Stevens

Beginner2022-09-24Added 12 answers

Maximum height occurs at x = b 2 a
From equation: we have;
a = 32 400 ,   b = 1 x = 1 32 400 = 400 32 x = 12.5  feet

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