A golf ball is hit off the top of a cliff that is 75 feet tall at an angle of 45° to the horizontal with an initial velocity of 80 feet per second. The quadratic equation shown below models the height, h(x), of the ball when it is x feet from the cliff’s edge. How far will the ball travel until it hits the ground? Round your answer to the nearest hundredth of a second.
Horizontal speed constant forever = 80 cos 45 = u = 56.6 ft/s
Initial vertical speed Vi = 80 sin 45 = 56.6 ft/s
Now you did not give me the quadratic so I will have to solve it with physics
v = Vi - g t
where g = about 32 ft/s^2 in these obsolete units
then
h = Hi + Vi t - 16 t^2
so
0 = 75 + 56.6 t - 16 t^2
but d = u t = 56.6 t
so t = d/56.6
and
0 = 75 + d - 16 (d^2/56.6^2)
0 = 75 + d - .005 d^2
.005 d^2 - d - 75 = 0
d^2 - 200 d - 15,016 = 0
d = [ 200 +/- sqrt ( 40,000+60,067)]/2
d = [ 200 +/- 316 ]/2
d = 258 feet
(ignore the negative distance, that is where it would have had to start if there were no cliff :)
To find out how far the ball will travel until it hits the ground, we need to determine the value of x when the height of the ball, h(x), equals 0. In other words, we want to find the x-intercepts of the quadratic equation.
The quadratic equation that models the height of the ball is h(x) = -16x^2 + 80x + 75, where x is the distance from the cliff's edge in feet.
To solve this equation, we set h(x) to 0 and solve for x:
-16x^2 + 80x + 75 = 0
This is a quadratic equation in the form ax^2 + bx + c = 0, where a = -16, b = 80, and c = 75. We can solve this equation by using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
x = (-80 ± √(80^2 - 4*(-16)*(75))) / (2*(-16))
x = (-80 ± √(6400 + 4800)) / (-32)
x = (-80 ± √11200) / (-32)
At this point, we need to take the positive square root only because we are looking for a distance:
x = (-80 + √11200) / (-32)
Using a calculator, we find that √11200 is approximately equal to 105.83.
x = (-80 + 105.83) / (-32)
x ≈ 0.93
Therefore, the ball will travel approximately 0.93 feet until it hits the ground.
To find out how far the ball will travel until it hits the ground, we need to find the value of x when the height of the ball is 0.
The quadratic equation model for the height, h(x), of the ball is given as follows:
h(x) = -16x^2 + x + 75
Setting h(x) to 0, we have:
-16x^2 + x + 75 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 1, and c = 75. Plugging these values into the quadratic formula, we get:
x = (-1 ± √(1^2 - 4(-16)(75))) / (2(-16))
Simplifying the equation further, we have:
x = (-1 ± √(1 + 4(16)(75))) / (-32)
x = (-1 ± √(1 + 4(1200))) / (-32)
x = (-1 ± √(1 + 4800)) / (-32)
x = (-1 ± √(4801)) / (-32)
Now, we can calculate the two possible values of x:
x = (-1 + √(4801)) / (-32) ≈ 18.07
x = (-1 - √(4801)) / (-32) ≈ -2.09
Since distance cannot be negative, the ball will travel approximately 18.07 feet until it hits the ground.
Therefore, the ball will travel approximately 18.07 feet from the cliff's edge until it hits the ground.