Question 824491: Solve the problem.
A projectile is fired from a cliff 100 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 170 feet per second. The height h of the projectile above the water is given by h(x)= -32x^2/(170)^2 + x + 100, where x is the horizontal distance of the projectile from the base of the cliff. How far from the base of the cliff is the height of the projectile a maximum?
Possible answers: A) 225.78ft B)451.56ft c)325.78ft D)777.34ft
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
h(x)= -32x^2/(170)^2 + x + 100
h(x)= -1.1072664e-3x^2 + x + 100
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the above quadratic equation is in standard form, with a=-1.1072664e-3, b=1, and c=100
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
-1.1072664e-3 1 100
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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this quadratic has two real roots (the x-intercepts), which are:
x = -90.8590996
x = 993.984129
the projectile strikes the ground 994 feet from the launch site.
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the quadratic vertex is a maximum at: ( x= 451.562515, h= 325.781257 )
the projectile reaches a maximum height of 325.8 feet above the ground
at a horizontal distance of 451.6 feet from the launch site
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