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 are given the equation 𝑦 = −5𝑥2 + 20𝑥 + 0.05, which represents the height of the golf ball in meters as a function of time in seconds. To determine how long the ball is in the air, we need to find the time when the height of the ball is zero (or equivalently, when it hits the ground).
Setting 𝑦 = 0 in the quadratic equation, we get:
0 = −5𝑥2 + 20𝑥 + 0.05
Using the quadratic formula, we can solve for 𝑥:
𝑥 = [−20 ± sqrt(20^2 − 4(−5)(0.05))] / (2(−5))
𝑥 = [−20 ± sqrt(400 + 1)] / (−10)
𝑥 = [−20 ± sqrt(401)] / (−10)
We get two solutions for 𝑥, corresponding to the two times when the ball is at zero height:
𝑥 = [−20 + sqrt(401)] / (−10) ≈ 4.03 seconds
𝑥 = [−20 - sqrt(401)] / (−10) ≈ 0.02 seconds
Since the second solution is negative, it doesn't make physical sense in this context. Therefore, we can conclude that the ball is in the air for approximately 4.03 seconds.
To determine how long the ball is in the air, we need to find the values of 𝑥 when 𝑦 = 0. The ball is in the air until it hits the ground, so when 𝑦 = 0, the ball has landed.
The quadratic equation 𝑦 = −5𝑥^2 + 20𝑥 + 0.05 can be rewritten as:
−5𝑥^2 + 20𝑥 + 0.05 = 0
To solve this quadratic equation, we can use the quadratic formula:
𝑥 = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐)) / (2𝑎)
In this case:
𝑎 = −5
𝑏 = 20
𝑐 = 0.05
Now we can substitute these values into the quadratic formula:
𝑥 = (−20 ± √(20^2 − 4(−5)(0.05))) / (2(−5))
Simplifying further:
𝑥 = (−20 ± √(400 + 1)) / (−10)
𝑥 = (−20 ± √401) / (−10)
Therefore, the ball is in the air for two time periods, which can be found by simplifying further:
𝑥 = (−20 + √401) / (−10)
AND
𝑥 = (−20 - √401) / (−10)
Calculating these values will give us the duration of the ball being in the air.