Question
Fireworks for the annual
4th of July show are launched straight up from a steel platform. The entire show is computer controlled. The height of a particular firework in meters above ground level is given by h=-4.9t^2+50t+11
, where time,
t , is in seconds.
a. What is the height of the platform?
b. How many seconds does it take for the firework to hit the ground?
c. What is the maximum height of the firework?
4th of July show are launched straight up from a steel platform. The entire show is computer controlled. The height of a particular firework in meters above ground level is given by h=-4.9t^2+50t+11
, where time,
t , is in seconds.
a. What is the height of the platform?
b. How many seconds does it take for the firework to hit the ground?
c. What is the maximum height of the firework?
Answers
GPT-4o mini
To analyze the motion of the firework given the height function \( h(t) = -4.9t^2 + 50t + 11 \), we will answer each question step-by-step.
### a. What is the height of the platform?
The height of the platform corresponds to the initial height of the firework when \( t = 0 \):
\[
h(0) = -4.9(0)^2 + 50(0) + 11 = 11 \, \text{meters}
\]
Thus, the height of the platform is **11 meters**.
---
### b. How many seconds does it take for the firework to hit the ground?
To find the time it takes for the firework to hit the ground, we need to determine when the height \( h(t) \) is zero:
\[
-4.9t^2 + 50t + 11 = 0
\]
This is a quadratic equation in the standard form \( at^2 + bt + c = 0 \). We can identify:
- \( a = -4.9 \)
- \( b = 50 \)
- \( c = 11 \)
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
1. Compute the discriminant \( b^2 - 4ac \):
\[
b^2 - 4ac = 50^2 - 4(-4.9)(11) = 2500 + 215.6 = 2715.6
\]
2. Now, plug the values into the quadratic formula:
\[
t = \frac{-50 \pm \sqrt{2715.6}}{2(-4.9)}
\]
First, calculate \( \sqrt{2715.6} \):
\[
\sqrt{2715.6} \approx 52.0
\]
3. Thus,
\[
t = \frac{-50 \pm 52.0}{-9.8}
\]
Calculating the two potential solutions:
- Positive root:
\[
t = \frac{-50 + 52.0}{-9.8} = \frac{2.0}{-9.8} \approx -0.204 \, \text{(not valid since time cannot be negative)}
\]
- Negative root:
\[
t = \frac{-50 - 52.0}{-9.8} = \frac{-102.0}{-9.8} \approx 10.41 \, \text{seconds}
\]
Therefore, it takes approximately **10.41 seconds** for the firework to hit the ground.
---
### c. What is the maximum height of the firework?
The maximum height of a projectile occurs at the vertex of the parabola defined by the quadratic equation. The time at which this maximum occurs can be found using the formula \( t = -\frac{b}{2a} \):
\[
t = -\frac{50}{2 \cdot -4.9} = \frac{50}{9.8} \approx 5.10 \, \text{seconds}
\]
Now, substituting \( t = 5.10 \) back into the height equation to find the maximum height:
\[
h(5.10) = -4.9(5.10)^2 + 50(5.10) + 11
\]
Calculating this step-by-step:
1. Calculate \( (5.10)^2 = 26.01 \).
2. Now plug this value in:
\[
h(5.10) = -4.9(26.01) + 50(5.10) + 11 = -127.649 + 255 + 11
\]
3. Summing it up:
\[
h(5.10) \approx -127.649 + 266 = 138.351 \, \text{meters}
\]
Thus, the maximum height of the firework is approximately **138.35 meters**.
### a. What is the height of the platform?
The height of the platform corresponds to the initial height of the firework when \( t = 0 \):
\[
h(0) = -4.9(0)^2 + 50(0) + 11 = 11 \, \text{meters}
\]
Thus, the height of the platform is **11 meters**.
---
### b. How many seconds does it take for the firework to hit the ground?
To find the time it takes for the firework to hit the ground, we need to determine when the height \( h(t) \) is zero:
\[
-4.9t^2 + 50t + 11 = 0
\]
This is a quadratic equation in the standard form \( at^2 + bt + c = 0 \). We can identify:
- \( a = -4.9 \)
- \( b = 50 \)
- \( c = 11 \)
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
1. Compute the discriminant \( b^2 - 4ac \):
\[
b^2 - 4ac = 50^2 - 4(-4.9)(11) = 2500 + 215.6 = 2715.6
\]
2. Now, plug the values into the quadratic formula:
\[
t = \frac{-50 \pm \sqrt{2715.6}}{2(-4.9)}
\]
First, calculate \( \sqrt{2715.6} \):
\[
\sqrt{2715.6} \approx 52.0
\]
3. Thus,
\[
t = \frac{-50 \pm 52.0}{-9.8}
\]
Calculating the two potential solutions:
- Positive root:
\[
t = \frac{-50 + 52.0}{-9.8} = \frac{2.0}{-9.8} \approx -0.204 \, \text{(not valid since time cannot be negative)}
\]
- Negative root:
\[
t = \frac{-50 - 52.0}{-9.8} = \frac{-102.0}{-9.8} \approx 10.41 \, \text{seconds}
\]
Therefore, it takes approximately **10.41 seconds** for the firework to hit the ground.
---
### c. What is the maximum height of the firework?
The maximum height of a projectile occurs at the vertex of the parabola defined by the quadratic equation. The time at which this maximum occurs can be found using the formula \( t = -\frac{b}{2a} \):
\[
t = -\frac{50}{2 \cdot -4.9} = \frac{50}{9.8} \approx 5.10 \, \text{seconds}
\]
Now, substituting \( t = 5.10 \) back into the height equation to find the maximum height:
\[
h(5.10) = -4.9(5.10)^2 + 50(5.10) + 11
\]
Calculating this step-by-step:
1. Calculate \( (5.10)^2 = 26.01 \).
2. Now plug this value in:
\[
h(5.10) = -4.9(26.01) + 50(5.10) + 11 = -127.649 + 255 + 11
\]
3. Summing it up:
\[
h(5.10) \approx -127.649 + 266 = 138.351 \, \text{meters}
\]
Thus, the maximum height of the firework is approximately **138.35 meters**.