A golf ball is hit from the top of a tee. The quadratic equation
𝑦 = −5𝑥2 + 20𝑥 + 0.05 describes its height, 𝑦, in metres as time, 𝑥, in seconds
passes. Determine how long the ball is in the air Use the quadratic formula.
We want to find the time when the height, 𝑦, is zero, since this would indicate that the golf ball has landed on the ground.
Setting 𝑦 equal to zero in the given equation, we get:
0 = −5𝑥2 + 20𝑥 + 0.05
Now we can use the quadratic formula:
𝑥 = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐))/2𝑎
Where 𝑎 = −5, 𝑏 = 20, and 𝑐 = 0.05.
Plugging in these values, we get:
𝑥 = (−20 ± √(20^2 − 4(−5)(0.05)))/2(−5)
Simplifying:
𝑥 = (−20 ± √(400.25))/−10
𝑥 ≈ 0.005 seconds or 𝑥 ≈ 3.995 seconds
The negative value doesn't make sense in this context, so we can discard it. Therefore, the golf ball is in the air for approximately 3.995 seconds.
To determine how long the ball is in the air, we need to find the values of 𝑥 when 𝑦 = 0 in the quadratic equation 𝑦 = −5𝑥^2 + 20𝑥 + 0.05.
Setting 𝑦 equal to 0, we have:
0 = −5𝑥^2 + 20𝑥 + 0.05
This equation is in the form 𝑎𝑥^2 + 𝑏𝑥 + 𝑐 = 0, where 𝑎 = -5, 𝑏 = 20, and 𝑐 = 0.05.
Using the quadratic formula:
𝑥 = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐)) / (2𝑎)
Substituting the values of 𝑎, 𝑏, and 𝑐 into the formula:
𝑥 = (−20 ± √(20^2 − 4(-5)(0.05))) / (2(-5))
Simplifying further:
𝑥 = (−20 ± √(400 + 1)) / (-10)
𝑥 = (−20 ± √401) / (-10)
Now, we can calculate the two possible values for 𝑥:
𝑥₁ = (−20 + √401) / (-10)
𝑥₂ = (−20 - √401) / (-10)
Using a calculator, we find:
𝑥₁ ≈ 3.955
𝑥₂ ≈ -0.455
Since time cannot be negative in this context, we can disregard 𝑥₂.
Therefore, the ball is in the air for approximately 3.955 seconds.