Question
A ball is thrown from a height of 44 meters with an initial downward velocity of 6 m/s. The ball's height h (in meters) after t seconds is given by the following.
H=44-6t-5t^2
How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
H=44-6t-5t^2
How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
Answers
GPT-4o mini
To find out when the ball hits the ground, we need to determine when the height \( h(t) = 0 \). The height of the ball is given by the equation:
\[
h(t) = 44 - 6t - 5t^2
\]
Setting this equal to zero to find when the ball hits the ground:
\[
44 - 6t - 5t^2 = 0
\]
Rearranging the equation gives us:
\[
-5t^2 - 6t + 44 = 0
\]
Multiplying the entire equation by -1 to make the leading coefficient positive:
\[
5t^2 + 6t - 44 = 0
\]
Now we will use the quadratic formula to solve for \( t \):
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In this case, \( a = 5 \), \( b = 6 \), and \( c = -44 \). We first calculate the discriminant:
\[
b^2 - 4ac = 6^2 - 4 \cdot 5 \cdot (-44) = 36 + 880 = 916
\]
Now using the quadratic formula:
\[
t = \frac{-6 \pm \sqrt{916}}{2 \cdot 5}
\]
Calculating \( \sqrt{916} \):
\[
\sqrt{916} \approx 30.233
\]
Now substituting back into the formula:
\[
t = \frac{-6 \pm 30.233}{10}
\]
Calculating the two possible values for \( t \):
1. Using the plus sign:
\[
t = \frac{-6 + 30.233}{10} = \frac{24.233}{10} \approx 2.4233
\]
2. Using the minus sign:
\[
t = \frac{-6 - 30.233}{10} = \frac{-36.233}{10} \approx -3.6233 \quad (\text{not physically meaningful, as time cannot be negative})
\]
Thus, the time when the ball hits the ground is:
\[
t \approx 2.42 \text{ seconds (rounded to the nearest hundredth)}
\]
The answer is:
**2.42** seconds.
\[
h(t) = 44 - 6t - 5t^2
\]
Setting this equal to zero to find when the ball hits the ground:
\[
44 - 6t - 5t^2 = 0
\]
Rearranging the equation gives us:
\[
-5t^2 - 6t + 44 = 0
\]
Multiplying the entire equation by -1 to make the leading coefficient positive:
\[
5t^2 + 6t - 44 = 0
\]
Now we will use the quadratic formula to solve for \( t \):
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In this case, \( a = 5 \), \( b = 6 \), and \( c = -44 \). We first calculate the discriminant:
\[
b^2 - 4ac = 6^2 - 4 \cdot 5 \cdot (-44) = 36 + 880 = 916
\]
Now using the quadratic formula:
\[
t = \frac{-6 \pm \sqrt{916}}{2 \cdot 5}
\]
Calculating \( \sqrt{916} \):
\[
\sqrt{916} \approx 30.233
\]
Now substituting back into the formula:
\[
t = \frac{-6 \pm 30.233}{10}
\]
Calculating the two possible values for \( t \):
1. Using the plus sign:
\[
t = \frac{-6 + 30.233}{10} = \frac{24.233}{10} \approx 2.4233
\]
2. Using the minus sign:
\[
t = \frac{-6 - 30.233}{10} = \frac{-36.233}{10} \approx -3.6233 \quad (\text{not physically meaningful, as time cannot be negative})
\]
Thus, the time when the ball hits the ground is:
\[
t \approx 2.42 \text{ seconds (rounded to the nearest hundredth)}
\]
The answer is:
**2.42** seconds.