An athlete whose event is the shot put releases the shot. When the shot is relea
ID: 3398382 • Letter: A
Question
An athlete whose event is the shot put releases the shot. When the shot is released at an angle of
25°,its path can be modeled by the formula y=-0.01x^2+0.5x+5.9 in which x is the shots horizontal distance, in feet, and y is its height, in feet. This formula is showb by one of the graphs, (a) or (b), in the figure. use the formula to answer the question below. QUESTION: Use the formula to determine the shots maximum distance. The distance is approximately ___ ft. (round to the nearest tenth as needed)
Explanation / Answer
f(x) = y = 6.1 + 0.6x - 0.01x²
A. The maximum height is the height at which dy/dx is zero
dy/dx = .6 - 0.02x = 0
x = 30
f(30) = 6.1 + 0.6*30 - .01*30² = 15.1
The maximum height is 15.1 feet and occurs 30 feet from the point of release.
B The height of the shot is zero when it hits the ground.
0 = 5.9 + 0.5x - 0.01x²
x = [-0.6±(0.6² + 4*0.01*6.1)]/(-0.01*2)
x = (-0.6±0.604)/(-0.02)
x (-0.6 ± 0.777)/(-0.02) (discard the negative result since the shot is traveling in the +x direction)
x 68.9 feet to the nearest tenth of a foot)
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