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A golf ball is hit so that it travels a horizontal distance of 440 feet and reaches a maximum height of 190 feet. Determine a quadratic equation that models the path of the golf ball, assuming it starts at the origin. (4 points) Round the coefficient of x2 to the nearest ten-thousandths place, and round the coefficient of x to the nearest thousandths place. Using your understanding of parametric equations for projectile motion, what is the angle the ball takes off? Show your steps.

Answer :

onyebuchinnaji

(1) A  quadratic equation that models the path of the golf ball, is 16.0850t² - 110.560t - 190 = 0.

(2) The angle the ball takes off is 64.4⁰.

Equation of motion of the golf ball

The equation that models the motion of the gofl ball is calculated as follows;

Vertical motion

Vf² = V₀² - 2gh

at maximum height, Vf = 0

V₀² = 2gh

V₀ = √2gh

V₀ = √(2 x 32.17 x 190)

V₀ = 110.56 ft/s

Equation of motion

h = V₀t - ¹/₂gt²

-190 = 110.56t  - ¹/₂(32.17)t²

-190 = 110.56t - 16.0850t²

16.0850t² - 110.560t - 190 = 0

Horizontal motion

440 = Vxt

Vx = 440/t

The time of motion, t is determined as follows;

16.0850t² - 110.560t + 190 = 0

a = 16.0850, b = -110.560, c = 190, solve using formula method

t = 8.3 s

Vx = 440 /8.3

Vx = 53.01 ft/s

Angle of projection

tan θ = (Vy) / (Vx)

tan θ = (110.56)/(53.01)

tan θ = 2.086

θ = tan⁻¹(2.086)

θ = 64.4⁰

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