A golf ball is hit with an initial velocity of 140 ft per second at an inclination of 45 degrees to the horizontal. In physics, it is established that the height h of the golf ball is given by the function h(x)=-32x^2/140^2+x,
How far does the golf ball travel before it lands?
Well, if a golf ball is hit with such velocity and inclination, you can bet it will go on quite an adventure! Now, to find out how far our little golf ball travels before it lands, we'll need to determine the value of x when h(x) (the height function) equals zero. Let's solve the equation:
-32x^2/140^2 + x = 0
Now, I may be a clown, but even I can't solve this algebraically. So, let's summon the help of our friend, Mr. Quadratic Formula:
x = [-b ± sqrt(b^2 - 4ac)] / (2a)
Plugging in the values of a, b, and c from our equation, we get:
x = [-(1) ± sqrt((1)^2 - 4(-32)(0)))] / (2(-32)/140^2)
Now, after doing some math (and clownish calculations), we end up with two values of x:
x = 0 or x = 9.858 ft approximately
Since we're interested in the positive value for x, we can say that the golf ball lands approximately 9.858 feet away. I hope it didn't cause too much havoc on its way down!
To find how far the golf ball travels before it lands, we need to determine the horizontal distance covered by the ball. This can be found by finding the value of x when the height h is equal to 0.
Given that the equation for the height of the golf ball is h(x) = -32x^2/140^2 + x, we set h(x) equal to 0 and solve for x:
0 = -32x^2/140^2 + x
To simplify the equation, let's multiply every term by 140^2 to get rid of the denominators:
0 = -32x^2 + 140^2x
Now, rewrite the equation in standard form:
32x^2 - 140^2x = 0
To solve this quadratic equation, we can factorize it:
x(32x - 140^2) = 0
Setting each factor equal to zero gives:
x = 0 or 32x - 140^2 = 0
Solving for x in the second equation:
32x - 140^2 = 0
32x = 140^2
x = (140^2)/32
Finally, calculate the value of x:
x ≈ 785.625
Therefore, the golf ball travels approximately 785.625 feet before it lands.
To determine how far the golf ball will travel before it lands, we need to find the horizontal distance traveled (denoted as x) when the vertical position (h) of the ball is zero.
Given that the equation for the height of the golf ball is h(x) = -32x^2/140^2 + x, we set it equal to zero to find the x-value at which the ball lands:
0 = -32x^2/140^2 + x
Since this equation is quadratic, let's first multiply through by 140^2 to get rid of the denominator:
0 = -32x^2 + 140^2x
Now, let's rearrange the equation to set it equal to zero and solve for x using the quadratic formula:
32x^2 - 140^2x = 0
Dividing by 4x, we obtain:
8x - 140^2 = 0
Now, let's use the quadratic formula, where a = 8, b = 0, and c = -140^2:
x = (-b ± √(b^2 - 4ac)) / (2a)
x = (± √(0^2 - 4(8)(-140^2))) / (2(8))
Simplifying further:
x = (± √(0 - 4(8)(-140^2))) / (16)
x = (± √(0 - (-4)(8)(140^2))) / (16)
x = (± √(4(8)(140^2))) / (16)
x = (± √(32(140^2))) / (16)
x = (± √(32) * √(140^2)) / 16
x = (± 4√2 * 140) / 16
x = (± 560√2) / 16
Reducing the fraction:
x = ± 35√2 / 2
Therefore, the golf ball will travel x = 35√2 / 2 feet before it lands.
Approximating the value:
Since √2 is roughly 1.414, we find:
x ≈ 35(1.414) / 2
x ≈ 49.37 feet
Therefore, the golf ball will travel approximately 49.37 feet before it lands.
just solve for x when h=0.
Clearly, it is only when x=0.
If you mean
h(x)=-32x^2/(140^2+x)
If you meant it as written,
h(x)=-32x^2/140^2 + x = x(1 - 64x/1225)
h=0 when x = 1225/64