SOLUTION: An athlete threw a ball from 2 ft. above the ground at a velocity of 100 feet per second (ft/s) and at an angle of 45 degress with respect to the ground. The path of the ball is gi
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Question 767080: An athlete threw a ball from 2 ft. above the ground at a velocity of 100 feet per second (ft/s) and at an angle of 45 degress with respect to the ground. The path of the ball is given by the function f(x) = -0.0016x^2+x+2, where f(x) is the height of the ball (in feet) and x is the distance traveled by the ball from the postition of the athlete. What is the maximum height reached by the ball? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! An athlete threw a ball from 2 ft. above the ground at a velocity of 100 feet per second (ft/s) and at an angle of 45 degress with respect to the ground. The path of the ball is given by the function f(x) = -0.0016x^2+x+2, where f(x) is the height of the ball (in feet) and x is the distance traveled by the ball from the postition of the athlete. What is the maximum height reached by the ball?
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f(x) = -0.0016x^2 + x + 2 is a parabola. Find the vertex.
The Line of Symmetry (LOS) is x = -b/2a
x = -1/(-0.0032) = 312.5
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Max ht is f(312.5)
= 158.25 feet
