The cup on the 9th hole of a golf course is located dead center in the middle of a circular green that is 70 ft in diameter. You ball is located 50 feet south, and 40 feet west of the cup. The ball follows a straight line path and exits the green at the right most edge. Assume the ball travels a constant rate of 10ft/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 up and when does this occur?
Assuming the "rightmost edge" of a circle is the end of the horizontal diameter, we have
hole is at H:(0,0)
ball is at B:(-40,-50)
ball exits at P:(35,0)
So, the line joining B and P is
y+50 = 2/3 (x+40)
The circle is represented by
x^2+y^2 = 35^2
The line intersects the circle at
Q:(-13,-32) and P:(35,0)
BQ = 32.5
So, assuming the ball is hit at time t, it reaches Q at t = 3.25
Now you can figure PQ and the time spent on the green.
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a. The ball enters the green in the southwestern direction, precisely 40 feet west and 50 feet south of the cup.
b. To calculate when the ball enters the green, we need to determine the distance it needs to travel to reach the edge of the green. Using the Pythagorean theorem, we can find the distance:
distance = √(40^2 + 50^2) = √(1600 + 2500) = √4100 ≈ 64.03 feet
The ball travels at a constant rate of 10 ft/sec, so to find the time it takes to enter the green, we divide the distance by the speed:
time = distance / speed = 64.03 ft / 10 ft/sec ≈ 6.4 seconds
Therefore, the ball enters the green approximately 6.4 seconds after it starts rolling.
c. The time the ball spends inside the green is equal to the time it takes to travel from the entrance point to the exit point. Since the exit point is on the rightmost edge of the green, the ball will need to travel the entire diameter of the green, which is 70 feet.
time = distance / speed = 70 ft / 10 ft/sec = 7 seconds
Thus, the ball spends approximately 7 seconds inside the green.
d. The ball is closest to the cup when it is directly west of it. Since the cup is located at the dead center of the circular green, it will be 35 feet west of the center. We can now calculate the distance between the ball and the cup:
distance = √(35^2 + 50^2) = √(1225 + 2500) = √3725 ≈ 61.09 feet
To find when the ball is closest to the cup, we need to determine the time it takes to travel this distance at a constant rate of 10 ft/sec:
time = distance / speed = 61.09 ft / 10 ft/sec ≈ 6.11 seconds
Hence, the ball is closest to the cup approximately 6.11 seconds after it starts rolling, when it is directly west of the cup.
a. To find where the ball enters the green, we can imagine a right-angled triangle formed by the ball, the cup, and the point where the ball enters the green. The legs of the triangle are 50 ft and 40 ft, representing the distances south and west from the ball to the cup respectively. Therefore, we can use the Pythagorean theorem to find the length of the hypotenuse, which represents the straight-line distance the ball travels before entering the green.
Using the Pythagorean theorem: c^2 = a^2 + b^2
where c is the length of the hypotenuse, a is the distance south (50 ft), and b is the distance west (40 ft).
c^2 = 50^2 + 40^2
c^2 = 2500 + 1600
c^2 = 4100
Taking the square root of both sides, we find:
c ≈ 64.03 ft
Therefore, the ball enters the green approximately 64.03 ft from the ball's initial position, which is 50 ft south and 40 ft west of the cup.
b. To find when the ball enters the green, we need to determine the time it takes for the ball to travel the straight-line distance of approximately 64.03 ft. We know that the ball travels at a constant rate of 10 ft/sec.
Using the formula: time = distance / rate
time = 64.03 ft / 10 ft/sec
time ≈ 6.403 sec
Therefore, the ball enters the green approximately 6.403 seconds after it starts moving.
c. To calculate the time spent inside the green, we need to determine the remaining time after the ball enters the green until it exits at the rightmost edge. The distance from the entry point to the exit point can be found by subtracting the radius of the green (35 ft) from the diameter (70 ft).
Distance from entry to exit = 70 ft - 35 ft = 35 ft
Using the formula: time = distance / rate
time = 35 ft / 10 ft/sec
time = 3.5 sec
Therefore, the ball spends approximately 3.5 seconds inside the green.
d. To find where the ball is located when it is closest to the cup and the corresponding time, we need to determine the point on the straight-line path where it is closest to the center of the cup. Since the ball exits the green at the rightmost edge, the closest point to the cup will be on the circular green.
We can determine the location and time by finding the point on the circumference of the green that forms a right angle with the line connecting the ball's entry point and the center of the cup.
The radius of the green is 35 ft, and we already know the entry point is 64.03 ft from the center of the cup. Using basic trigonometry, we can calculate the angle formed between the radius and the line connecting the entry point and the center of the cup.
sin(angle) = opposite / hypotenuse
sin(angle) = 35 / 64.03
angle ≈ 34.86 degrees
Since the angle is formed between the radius and the line to the center of the cup, we can double it to find the angle between the line connecting the entry point and the center of the cup.
Angle = 2 * 34.86
Angle ≈ 69.72 degrees
Now that we have the angle, we can calculate the distance from the entry point to the closest point on the circumference of the green.
Distance = radius * sin(angle)
Distance ≈ 35 * sin(69.72)
Distance ≈ 33.71 ft
Therefore, the ball is located approximately 33.71 ft from the center of the cup when it is closest.
To find the time, we can divide this distance by the ball's constant rate of 10 ft/sec.
Time = distance / rate
Time = 33.71 ft / 10 ft/sec
Time ≈ 3.371 sec
Therefore, the ball is closest to the cup approximately 3.371 seconds after it starts moving.