Question 816071: A projectile is fired from a cliff 400 feet above water at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 160 feet per second. The height h of the projectile above the water is given by h(x)=-32x^2/(160)^2+x+400, where x is the horizontal distance of the projectile from the base of the cliff. Find the maximum height of the projectile.
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! h(x) = -32x^2/(160)^2 + x + 400
-32x^2/(160)^2 + x + 400
-0.00125x^2 + x + 400
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the above quadratic equation is in standard form, with a=-0.00125, b=1, and c=400
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to solve the quadratic equation, plug this:
-0.00125 1 400
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic vertex is a maximum at ( x=400, h(x)=600 )
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Answer:
the projectile reaches a maximum height of 600 feet above the water
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