An athlete whose event is the shot put releases the shot Whe

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)

Solution

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)