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 Hint: Use the quadratic formula.
AAAaannndd the bot gets it wrong yet again!
The ball is in the air until the height is zero. You correctly found the roots
x = (20 ± 2√101) / 10 = 4.0025 or -0.025
so the ball was in the air for 4.0025 seconds
Nobody asked anything about the vertex.
I apologize for my mistake in the previous response. You are correct that the ball is in the air until the height is zero, and the roots of the quadratic equation are the times when the height is zero. Using the quadratic formula:
𝑥 = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐)) / (2𝑎)
with 𝑎 = −5, 𝑏 = 20, and 𝑐 = 0.05, we get:
𝑥 = (−20 ± √(20^2 − 4(−5)(0.05))) / (2(−5))
which simplifies to:
𝑥 = (20 ± 2√101) / 10
Therefore, the ball is in the air for approximately 4.0025 seconds (using the positive root), not 0.05 seconds as previously stated. Thank you for bringing this to my attention, and I apologize for any confusion caused by my mistake.
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.
This can be done by using the quadratic formula, which is given by:
𝑥 = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐))/(2𝑎)
In our equation, 𝑎 = -5, 𝑏 = 20, and 𝑐 = 0.05. Plugging in these values into the formula, we get:
𝑥 = (−20 ± √(20^2 - 4(-5)(0.05)))/(2(-5))
Simplifying further:
𝑥 = (−20 ± √(400 + 1))/(−10)
𝑥 = (−20 ± √401)/(−10)
Now we have two possible solutions for 𝑥:
𝑥₁ = (−20 + √401)/(−10) --> This gives the greater value of 𝑥, representing the time the ball touches the ground.
𝑥₂ = (−20 - √401)/(−10) --> This gives the lesser value of 𝑥, which is not relevant for this problem.
Evaluating 𝑥₁ numerically:
𝑥₁ = (−20 + √401)/(−10)
𝑥₁ ≈ (−20 + 20.02)/(−10)
𝑥₁ ≈ (0.02)/(−10)
𝑥₁ ≈ -0.002
Since time cannot be negative, we can conclude that the ball is in the air for approximately 0.002 seconds.
To determine how long the ball is in the air, we need to find the roots of the quadratic equation 𝑦 = −5𝑥2 + 20𝑥 + 0.05, which represent the times when the height of the ball is 0 (i.e. when it hits the ground). We can use the quadratic formula:
𝑥 = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐)) / (2𝑎)
where 𝑎 = −5, 𝑏 = 20, and 𝑐 = 0.05.
Substituting these values into the formula, we get:
𝑥 = (−20 ± √(20^2 − 4(−5)(0.05))) / (2(−5))
𝑥 = (−20 ± √404) / (−10)
Simplifying this expression, we get:
𝑥 = (20 ± 2√101) / 10
Therefore, the ball is in the air for the duration of time given by:
𝑥 = (20 + 2√101) / 10 ≈ 2.05 seconds
or
𝑥 = (20 − 2√101) / 10 ≈ 0.05 seconds
Since the quadratic equation has a negative coefficient for the quadratic term, the graph is a downward-facing parabola. This means that the ball reaches its maximum height at the vertex of the parabola, which occurs at a time of:
𝑥 = −𝑏 / (2𝑎) = −20 / (2(−5)) = 2 seconds
Therefore, the ball is in the air for 2.05 − 2 = 0.05 seconds.