Jack talks to water balloon into the air from the top of his shed in the backyard graph the situation using the formula y equals negative 16 squared + 30t + 8 when will the balloon hit the ground round your answer to the nearest tenth of a second

Answer 1

To determine when the water balloon will hit the ground, we can use the formula provided:

\[ y = -16t^2 + 30t + 8 \]

Here, \( y \) represents the height of the balloon (in feet), and \( t \) represents time (in seconds). The balloon will hit the ground when \( y = 0 \). Therefore, we need to solve the equation:

\[ -16t^2 + 30t + 8 = 0 \]

To solve this quadratic equation, we can use the quadratic formula, which is given by:

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In this case, \( a = -16 \), \( b = 30 \), and \( c = 8 \). Plugging these values into the formula gives:

1. Calculate \( b^2 - 4ac \):

\[ b^2 - 4ac = 30^2 - 4(-16)(8) \]

\[ = 900 + 512 \]

\[ = 1412 \]

2. Now use the quadratic formula:

\[ t = \frac{-30 \pm \sqrt{1412}}{2(-16)} \]

First, calculate \( \sqrt{1412} \):

\[ \sqrt{1412} \approx 37.5 \]

Now substitute that back in:

\[ t = \frac{-30 \pm 37.5}{-32} \]

This gives us two potential solutions for \( t \):

\[ t = \frac{-30 + 37.5}{-32} \quad \text{and} \quad t = \frac{-30 - 37.5}{-32} \]

Calculating each:

1. For \( -30 + 37.5 \):

\[ t = \frac{7.5}{-32} \approx -0.234375 \quad (\text{not valid as time cannot be negative}) \]

2. For \( -30 - 37.5 \):

\[ t = \frac{-67.5}{-32} \approx 2.109375 \]

Rounding this to the nearest tenth, we find:

\[ \boxed{2.1} \text{ seconds} \]

Thus, the water balloon will hit the ground approximately 2.1 seconds after being thrown.