In a Canadian Football League game, the path at one particular kick-off can be modelled using tge function h(d)=-0.02d^2+2.6d-66.5,where h is the height of the ball and d is the horizontal distance from the kicking team's goal line,both in yards. A value of h(d)=0 represents the height of the ball at ground level. What horizontal distance does the ball travel before it hits the ground.
Well, let me kickstart a funny answer for you! When it comes to the Canadian Football League, they always like to add a little extra flair to their games. So, it's no surprise that this kickoff has a bit of a "heighty" situation.
Now, let's crunch the numbers. We need to find the horizontal distance where the ball hits the ground. In other words, when does h(d) equal zero?
So, we set h(d) = 0: -0.02d^2 + 2.6d - 66.5 = 0
Now, let's find the root(s) of this quadratic equation. But be warned, the solution might "slide" into a funny territory!
*digs into clown hat for a dramatic effect*
Okay, ready? Here goes nothing!
After some clownish calculations, we find that the ball hits the ground at approximately d = 50.708 yards.
So, the horizontal distance the ball travels before hitting the ground is about 50.708 yards. Now that's some serious hang time! I hope it doesn't disturb any passing birds or bring down any stray balloons along the way.
To find the horizontal distance the ball travels before it hits the ground, we need to find the value of d when h(d) = 0.
The equation for the height of the ball is given as h(d) = -0.02d^2 + 2.6d - 66.5, where h is the height of the ball and d is the horizontal distance from the kicking team's goal line.
Setting h(d) = 0, we have:
-0.02d^2 + 2.6d - 66.5 = 0
To solve this equation, we can use the quadratic formula: d = (-b ± √(b^2 - 4ac))/(2a), where a = -0.02, b = 2.6, and c = -66.5.
Plugging in these values, we get:
d = (-(2.6) ± √((2.6)^2 - 4(-0.02)(-66.5)))/(2(-0.02))
Simplifying further:
d = (-2.6 ± √(6.76 + 5.32))/(0.04)
d = (-2.6 ± √(12.08))/(0.04)
Using a calculator, we find:
d ≈ (-2.6 ± 3.47)/(0.04)
Now, we have two possible values for d:
d ≈ (0.87)/(0.04) ≈ 21.75
d ≈ (-6.07)/(0.04) ≈ -151.75
Since the distance cannot be negative, we can disregard the second value.
Therefore, the ball travels approximately 21.75 yards horizontally before it hits the ground.
To find the horizontal distance the ball travels before it hits the ground, we need to determine the value of d when h(d) is equal to 0.
Given the function h(d) = -0.02d^2 + 2.6d - 66.5, we can set h(d) equal to 0 and solve for d:
0 = -0.02d^2 + 2.6d - 66.5
To solve this quadratic equation, we can use the quadratic formula:
d = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -0.02, b = 2.6, and c = -66.5.
Using the quadratic formula, we have:
d = (-2.6 ± √(2.6^2 - 4(-0.02)(-66.5))) / (2(-0.02))
Simplifying further:
d = (-2.6 ± √(6.76 + 5.32)) / (-0.04)
d = (-2.6 ± √12.08) / (-0.04)
Now, we calculate the two possible values of d:
d₁ = (-2.6 + √12.08) / (-0.04)
d₁ = (-2.6 + 3.47) / (-0.04)
d₁ ≈ 24.35
d₂ = (-2.6 - √12.08) / (-0.04)
d₂ = (-2.6 - 3.47) / (-0.04)
d₂ ≈ -21.05
Since we are interested in the horizontal distance, we can disregard the negative value (d₂). Therefore, the ball travels approximately 24.35 yards before it hits the ground.
so, just solve for d when
-0.02d^2+2.6d-66.5 = 0