Question
A ball is thrown from a height of 217 feet with an Initial downward velocity of 17 ft/s. The ball's height h (in feet) after t seconds is given by the following.
H=217-17t-16t^2
How long after the ball is thrown does it hit the ground?
Round your answers) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
H=217-17t-16t^2
How long after the ball is thrown does it hit the ground?
Round your answers) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
Answers
GPT-4o mini
To find how long after the ball is thrown it hits the ground, we need to solve the equation for \( h = 0 \):
\[
217 - 17t - 16t^2 = 0
\]
Rearranging this gives us:
\[
-16t^2 - 17t + 217 = 0
\]
Multiplying the entire equation by -1 to make it standard form:
\[
16t^2 + 17t - 217 = 0
\]
Now, we can use the quadratic formula to solve for \( t \). The quadratic formula is given by:
\[
t = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}
\]
In our case, \( a = 16 \), \( b = 17 \), and \( c = -217 \). First, we need to calculate the discriminant \( b^2 - 4ac \):
\[
b^2 - 4ac = 17^2 - 4 \cdot 16 \cdot (-217)
\]
Calculating \( b^2 \):
\[
17^2 = 289
\]
Calculating \( 4ac \):
\[
4 \cdot 16 \cdot (-217) = -13952
\]
Now putting it together:
\[
b^2 - 4ac = 289 + 13952 = 14241
\]
Now we can find \( \sqrt{14241} \):
\[
\sqrt{14241} \approx 119.34 \quad (\text{using a calculator})
\]
Putting it back into the quadratic formula:
\[
t = \frac{{-17 \pm 119.34}}{{2 \cdot 16}}
\]
Calculating the two potential solutions for \( t \):
1. First solution:
\[
t = \frac{{-17 + 119.34}}{{32}} = \frac{{102.34}}{{32}} \approx 3.20
\]
2. Second solution:
\[
t = \frac{{-17 - 119.34}}{{32}} = \frac{{-136.34}}{{32}} \approx -4.26
\]
Since time cannot be negative, the solution that makes sense is:
\[
t \approx 3.20 \text{ seconds}
\]
Thus, the ball hits the ground approximately **3.20 seconds** after it was thrown.
\[
217 - 17t - 16t^2 = 0
\]
Rearranging this gives us:
\[
-16t^2 - 17t + 217 = 0
\]
Multiplying the entire equation by -1 to make it standard form:
\[
16t^2 + 17t - 217 = 0
\]
Now, we can use the quadratic formula to solve for \( t \). The quadratic formula is given by:
\[
t = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}
\]
In our case, \( a = 16 \), \( b = 17 \), and \( c = -217 \). First, we need to calculate the discriminant \( b^2 - 4ac \):
\[
b^2 - 4ac = 17^2 - 4 \cdot 16 \cdot (-217)
\]
Calculating \( b^2 \):
\[
17^2 = 289
\]
Calculating \( 4ac \):
\[
4 \cdot 16 \cdot (-217) = -13952
\]
Now putting it together:
\[
b^2 - 4ac = 289 + 13952 = 14241
\]
Now we can find \( \sqrt{14241} \):
\[
\sqrt{14241} \approx 119.34 \quad (\text{using a calculator})
\]
Putting it back into the quadratic formula:
\[
t = \frac{{-17 \pm 119.34}}{{2 \cdot 16}}
\]
Calculating the two potential solutions for \( t \):
1. First solution:
\[
t = \frac{{-17 + 119.34}}{{32}} = \frac{{102.34}}{{32}} \approx 3.20
\]
2. Second solution:
\[
t = \frac{{-17 - 119.34}}{{32}} = \frac{{-136.34}}{{32}} \approx -4.26
\]
Since time cannot be negative, the solution that makes sense is:
\[
t \approx 3.20 \text{ seconds}
\]
Thus, the ball hits the ground approximately **3.20 seconds** after it was thrown.