Question
Problem 2 Scenario:(3pts)
A ball is thrown from a height of 5 feet with an initial velocity of 48 feet per second at an angle of 45 degrees above the horizontal. The parametric equations are:
xt=24t yt= -16t2+24t+5
How long does it take for the ball to reach a horizontal distance of 36 feet?
At what time does the ball hit the ground?
(Note: Use the quadratic formula to solve.)
A ball is thrown from a height of 5 feet with an initial velocity of 48 feet per second at an angle of 45 degrees above the horizontal. The parametric equations are:
xt=24t yt= -16t2+24t+5
How long does it take for the ball to reach a horizontal distance of 36 feet?
At what time does the ball hit the ground?
(Note: Use the quadratic formula to solve.)
Answers
GPT-4o mini
The given parametric equations for the ball's motion are:
\[
x(t) = 24t
\]
\[
y(t) = -16t^2 + 24t + 5
\]
### Part 1: Time to reach a horizontal distance of 36 feet
To find the time \(t\) when the ball reaches a horizontal distance of 36 feet, we set the equation for \(x(t)\) equal to 36:
\[
24t = 36
\]
Now, solve for \(t\):
\[
t = \frac{36}{24} = \frac{3}{2} = 1.5
\]
### Part 2: Time when the ball hits the ground
To find the time \(t\) when the ball hits the ground, we need to find when \(y(t) = 0\):
\[
-16t^2 + 24t + 5 = 0
\]
This is a quadratic equation in standard form \(at^2 + bt + c = 0\), where:
- \(a = -16\)
- \(b = 24\)
- \(c = 5\)
We can use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substituting in the values for \(a\), \(b\), and \(c\):
\[
t = \frac{-24 \pm \sqrt{24^2 - 4(-16)(5)}}{2(-16)}
\]
Now, calculate the discriminant:
\[
24^2 = 576
\]
\[
-4 \cdot -16 \cdot 5 = 320
\]
\[
b^2 - 4ac = 576 + 320 = 896
\]
Now, substitute back into the quadratic formula:
\[
t = \frac{-24 \pm \sqrt{896}}{-32}
\]
Next, calculate \(\sqrt{896}\):
\[
\sqrt{896} = \sqrt{16 \cdot 56} = 4\sqrt{56} = 4 \cdot \sqrt{4 \cdot 14} = 8\sqrt{14}
\]
Substituting this back into the formula:
\[
t = \frac{-24 \pm 8\sqrt{14}}{-32}
\]
\[
t = \frac{24 \mp 8\sqrt{14}}{32}
\]
\[
t = \frac{3 \mp \sqrt{14}}{4}
\]
Since we are looking for the time when the ball hits the ground, we will take the positive solution:
\[
t = \frac{3 + \sqrt{14}}{4}
\]
### Summary
1. Time to reach horizontal distance of 36 feet: \( t = 1.5 \) seconds.
2. Time when the ball hits the ground: \( t = \frac{3 + \sqrt{14}}{4} \) seconds, approximately \( t \approx 1.85 \) seconds (since \( \sqrt{14} \approx 3.74 \)).
\[
x(t) = 24t
\]
\[
y(t) = -16t^2 + 24t + 5
\]
### Part 1: Time to reach a horizontal distance of 36 feet
To find the time \(t\) when the ball reaches a horizontal distance of 36 feet, we set the equation for \(x(t)\) equal to 36:
\[
24t = 36
\]
Now, solve for \(t\):
\[
t = \frac{36}{24} = \frac{3}{2} = 1.5
\]
### Part 2: Time when the ball hits the ground
To find the time \(t\) when the ball hits the ground, we need to find when \(y(t) = 0\):
\[
-16t^2 + 24t + 5 = 0
\]
This is a quadratic equation in standard form \(at^2 + bt + c = 0\), where:
- \(a = -16\)
- \(b = 24\)
- \(c = 5\)
We can use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substituting in the values for \(a\), \(b\), and \(c\):
\[
t = \frac{-24 \pm \sqrt{24^2 - 4(-16)(5)}}{2(-16)}
\]
Now, calculate the discriminant:
\[
24^2 = 576
\]
\[
-4 \cdot -16 \cdot 5 = 320
\]
\[
b^2 - 4ac = 576 + 320 = 896
\]
Now, substitute back into the quadratic formula:
\[
t = \frac{-24 \pm \sqrt{896}}{-32}
\]
Next, calculate \(\sqrt{896}\):
\[
\sqrt{896} = \sqrt{16 \cdot 56} = 4\sqrt{56} = 4 \cdot \sqrt{4 \cdot 14} = 8\sqrt{14}
\]
Substituting this back into the formula:
\[
t = \frac{-24 \pm 8\sqrt{14}}{-32}
\]
\[
t = \frac{24 \mp 8\sqrt{14}}{32}
\]
\[
t = \frac{3 \mp \sqrt{14}}{4}
\]
Since we are looking for the time when the ball hits the ground, we will take the positive solution:
\[
t = \frac{3 + \sqrt{14}}{4}
\]
### Summary
1. Time to reach horizontal distance of 36 feet: \( t = 1.5 \) seconds.
2. Time when the ball hits the ground: \( t = \frac{3 + \sqrt{14}}{4} \) seconds, approximately \( t \approx 1.85 \) seconds (since \( \sqrt{14} \approx 3.74 \)).