### College Pranks: Calculating the Time for a Water Balloon to Hit the GroundA student uses rubber tubing to launch a water balloon from the roof of his dormitory. The height $ h $ (in feet) of the balloon, $ t $ seconds after being launched, is given by the formula \[ h = -16t^2 + 48t + 64. \] After how many seconds will the balloon hit the ground?To find out when the water balloon hits the ground, we need to determine when the height $ h $ will be 0 feet. So, we set $ h $ equal to 0 and solve the quadratic equation for $ t $.#### Step-by-Step Solution1. **Set the height to 0** (because the balloon hits the ground when its height is 0): \[ h = -16t^2 + 48t + 64 \] \[ 0 = -16t^2 + 48t + 64 \]2. **Factor the quadratic equation**: \[ 0 = -16(t^2 - 3t - 4) \] \[ 0 = -16(t + 1)(t - 4) \] - Factor out the opposite of the Greatest Common Factor (GCF), which is -16. - Factor the trinomial $ t^2 - 3t - 4 $.3. **Solve for $ t $** by setting each factor that contains a variable to 0: \[ t + 1 = 0 \quad \text{or} \quad t - 4 = 0 \] \[ t = -1 \quad \text{or} \quad t = 4 \]4. **Interpret the solutions**: - Discard $ t = -1 $ because time cannot be negative. - The valid solution is $ t = 4 $, indicating that the balloon hits the ground 4 seconds after being launched.#### VerificationCheck this result by substituting $ t = 4 $ into the original height equation:\[ h = -16(4)^2 + 48(4) + 64 \]\[ h = -16(16) + 192 + 64 \]\[ h = -256 + 192 + 64 \]\[ h =

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College Pranks. A student uses rubber tubing to launch a water balloon from the roof of his dormitory. The height h (in feet) of the balloon, t seconds after being launched, is given by the formula h = -16t 2 + 48t + 64. After how many seconds will the balloon hit the ground? When the water balloon hits the ground, its height will be 0 feet.To find the time that it takes for the balloon to hit the ground, we set h equal to 0, and solve the quadratic equation for t. h = -16t 2 + 48t + 64 0 = -16t 2 + 48t + 64 Substitute 0 for the height, h. This is a quadratic equation. 0 = -16t 2 + 48t + 64 0 = -16(t2 - 3t - 0 = -16(t + Factor out the opposite of the GCF, -16. )(t – 4) Factor the trinomial. t -[ | = 0 Since -16 cannot equal 0, discard that possibility. Set each factor that contains a variable equal to 0. t + = 0 or 1 = -1 t = 4 Solve each equation. The equation has two solutions, -1 and 4. Since t represents time, and, in this case, time cannot be negative, we discard -1. The second solution, 4, indicates that the balloon hits the ground 4 second after being launched. Check this result by substituting 4 for t in h = -16t 2 + 48t + 64. You should get h =

College Pranks. A student uses rubber tubing to launch a water balloon from the roof of his dormitory. The height h (in feet) of the balloon, t seconds after being launched, is given by the formula h = -16t 2 + 48t + 64. After how many seconds will the balloon hit the ground? When the water balloon hits the ground, its height will be 0 feet.To find the time that it takes for the balloon to hit the ground, we set h equal to 0, and solve the quadratic equation for t. h = -16t 2 + 48t + 64 0 = -16t 2 + 48t + 64 Substitute 0 for the height, h. This is a quadratic equation. 0 = -16t 2 + 48t + 64 0 = -16(t2 - 3t - 0 = -16(t + Factor out the opposite of the GCF, -16. )(t – 4) Factor the trinomial. t -[ | = 0 Since -16 cannot equal 0, discard that possibility. Set each factor that contains a variable equal to 0. t + = 0 or 1 = -1 t = 4 Solve each equation. The equation has two solutions, -1 and 4. Since t represents time, and, in this case, time cannot be negative, we discard -1. The second solution, 4, indicates that the balloon hits the ground 4 second after being launched. Check this result by substituting 4 for t in h = -16t 2 + 48t + 64. You should get h =

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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### College Pranks: Calculating the Time for a Water Balloon to Hit the Ground

A student uses rubber tubing to launch a water balloon from the roof of his dormitory. The height $ h $ (in feet) of the balloon, $ t $ seconds after being launched, is given by the formula
\[ h = -16t^2 + 48t + 64. \]
After how many seconds will the balloon hit the ground?

To find out when the water balloon hits the ground, we need to determine when the height $ h $ will be 0 feet. So, we set $ h $ equal to 0 and solve the quadratic equation for $ t $.

#### Step-by-Step Solution

1. **Set the height to 0** (because the balloon hits the ground when its height is 0):
\[ h = -16t^2 + 48t + 64 \]
\[ 0 = -16t^2 + 48t + 64 \]

2. **Factor the quadratic equation**:
\[ 0 = -16(t^2 - 3t - 4) \]
\[ 0 = -16(t + 1)(t - 4) \]

- Factor out the opposite of the Greatest Common Factor (GCF), which is -16.
- Factor the trinomial $ t^2 - 3t - 4 $.

3. **Solve for $ t $** by setting each factor that contains a variable to 0:
\[ t + 1 = 0 \quad \text{or} \quad t - 4 = 0 \]
\[ t = -1 \quad \text{or} \quad t = 4 \]

4. **Interpret the solutions**:
- Discard $ t = -1 $ because time cannot be negative.
- The valid solution is $ t = 4 $, indicating that the balloon hits the ground 4 seconds after being launched.

#### Verification

Check this result by substituting $ t = 4 $ into the original height equation:
\[ h = -16(4)^2 + 48(4) + 64 \]
\[ h = -16(16) + 192 + 64 \]
\[ h = -256 + 192 + 64 \]
\[ h =

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