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
Well, I would write that as
y = 0.05 + 20 t - 5 t^2
the initial height when t = 0 is y = 0.05
the initial velocity up is 20
and the acceleration down is 2 * 5 = 10
so we usually write that as
h = Hi + Vi t - (1/2) g t^2
that said returning to the problem
5 t^2 - 20 t - 0.05 = 0
what is t when h = 0 standard old quadratic
t = [ 20 +/- sqrt (400 + 1) ] / 10
looks like about 4 seconds
By the way. 0.05 meters is a very low tree !! Typo?
It's possible that it's a low tee used for putting, or it could be a typo as you suggested. However, either way, our calculations are correct assuming that the initial height is 0.05 meters.
Oh, sorry, misread tree for tee :)
No problem! It happens to the best of us.
To determine how long the ball is in the air, we need to find the values of x where y is greater than zero (since the ball has positive height when it's in the air).
The quadratic equation y = -5x^2 + 20x + 0.05 can be set to zero to find the x-intercepts:
-5x^2 + 20x + 0.05 = 0
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Where a, b, and c are the coefficients of the quadratic equation. In our case, a = -5, b = 20, and c = 0.05. Plugging these values into the quadratic formula, we have:
x = (-20 ± √(20^2 - 4(-5)(0.05))) / 2(-5)
Simplifying further:
x = (-20 ± √(400 + 1)) / -10
x = (-20 ± √401) / -10
Now, let's calculate the values of x:
x1 = (-20 + √401) / -10
x2 = (-20 - √401) / -10
x1 ≈ 0.087
x2 ≈ 3.613
Since we are interested in the positive values of x, we discard x2 because it's negative.
Therefore, the ball is in the air for approximately 0.087 seconds.
To determine how long the golf ball is in the air, we need to find the values of x when the height y is equal to 0. This is because the ball is on the ground when it has a height of 0.
Let's set the equation y = -5x^2 + 20x + 0.05 equal to 0 and solve for x:
-5x^2 + 20x + 0.05 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -5, b = 20, and c = 0.05.
Plugging these values into the quadratic 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 values of x:
x1 = (-20 + √401) / (-10)
x2 = (-20 - √401) / (-10)
To determine how long the ball is in the air, we should consider only the positive value of x since time cannot be negative. Therefore, the ball is in the air for x = (-20 + √401) / (-10) seconds.