Written problem bullet A golfer is playing on a hill that is

Written problem bullet A golfer is playing on a hill that is described by the parabolic equation (ft): y= -(x-700/50)^2 +200bullet The ball is hits the ball at 80mph at a 13 degree angle with respect to the slope of the

Solution

Slope of the hill

m= dy/dx = -2(x-700)/50*(1/50)

                  = 14/25-x/1250

At x=0 slope = 14/25

Angle of projection = 13+arctan(14/25) = 42.25⁰

Initial velocity of projectile   u = 80mph = 117.33 ft/s

Projectile motion

x = uCos*t = 117.33*Cos(42.25)*t = 86.85t

y = uSin*t –gt² /2 = 117.33*Sin(42.25)*t-32t²/2

        = 78.89t – 16t²   (we take g= 32ft/s² )

We substitute the above values of x and y in the parabolic equation of the hill to know where the ball strikes the hill

78.89t – 16t² = -((86.85t-700)/50)² +200

12.98t² -30.25t +4   = 0

Solving the quadratic we get t = 0.14s, 2.19 s

We take the first value of t =0.14

X= 86.85 *0.14 = 12.21 ft

Y = 78.89*0.14 -16*0.14² = 10.73 ft

The ball strikes the hill

Land distance = sqrt(x² + y²) = sqrt(12.21² +10.73²) = 16.25 ft

Slope of the hill = m = 14/25-12.21/1250   = 0.55

            = 28.82⁰ with the horizontal

the ball makes an angle of arctan(10.73/12.21) = 41.31⁰ with the horizontal

angle of landing on the hill = 41.31 -28.82 = 12.49⁰

average velocity = 16.25/0.14 = 116.07 ft/s