The cup on the 9th hole of a golf course is located dead center in the middle of a circular green that is 70 feet in diameter. your ball is located as in the picture below: the ball follows a straight line path and exits the green at the right-most edge. assume the ball travels a constant rate of 10 ft/sec.
(a) where does the ball enter the green? (b) when does the ball enter the green? (c) how long does the ball spend inside the green? (d) where is the ball located when it is closest to the cup and when does this occur. the ball follows a straight

As the cup of the 9th hole in the golf course is located as the dead in the center of the milled green circular shape that is 70 feet. A ball follows the straight-line path and exists at the fright most edge.  

Learn more about the hole of a golf course is located dead center in the middle.

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a. The ball enters the green at Q (-13.46, -32.31)

b. 3.189s

c. 5.824s

d. Closest point =  (10.77, -16.15), it occurs 6.102s after the ball starts its journey

Step-by-step explanation:

he rightmost edge is taken as the right end of the horizontal diameter

Coordinates of the hole at H:(0,0)

Coordinates of the ball B:(-40,-50)

Coordinates of the point the ball exits green P:(35,0)

So, the line joining B and P is given by

Looking for the gradient of the line B P we have

(x2-x1)/(y1-y2) =(35 –(-40))/(0-(-50))

 = 75/50=3/2

The line equation therefore = (y-(-50))/(x-(-50)) = 2/3

(y + 50)/(x+40) =2/3

y+50 = 2/3 (x+40)

The equation of a circle is given by

x^2+y^2 = 35^2

 solving the two equations we have

x = ((3y+150)/2)-40 = 1.5y-35

Substituting the value of x into the above equations we have (1.5y-35)2 + y2 = 352

3.25×y2+105×y = 0

From where we have y×(3.25×y + 105) = 0

From where y = 0 or y = -105/3.25 = -32.31m

Also x = 35 or x = -13.46

The line intersects the circle at

Q:(-13.46,-32.31) and P:(35,0)

a. The ball enters the green at Q (-13.46, -32.31)

b. The ball enters the green after (length BQ)/Velocity

Where length BQ is given as

√((-13.46-(-40))2+(-32.31-(-50))2) = √(26.532+17.692)  = 31.89ft

The ball enters after 31.89ft/10ft/s or 3.189s

c. To find out how long the ball spends inside the green we have

Q:(-13.46,-32.31) and P:(35,0)

Length QP2 = (-13.46-(35))2 + (-32.31- 0)2 = 3392.31 = 58.24m

Time spent in the green = 58.24/10 = 5.824s

d. The ball is closest to the cup at the point when the perpendicular from the cup intersects the line PQ

and this is at the center of the line PQ hence the locus of the point is

((-13.46 + (35-(-13.46))/2), (-32.31+(0-(-32.31))/2) =  (10.77, -16.15)

This occurs at (√( (10.77-(-40))2 + (-16.15(-50))2))/10 = 61.02ft/10ft/s = 6.102s after the ball starts its journey

solve systems of equations by graphing. a system of linear equations contains two or more equations e.g. y=0.5x+2 and y=x-2. the solution of such a system is the ordered pair that is a solution to both equations. to solve a system of linear equations graphically we graph both equations in the same coordinate system

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step-by-step explanation: