Worksheet featuring four quadratic equation word problems related to physics and business applications.

Problem 1: Flare Launched from a Boat

The height of the flare above the water is given by the function:
\[ h(t) = -15t^2 + 150t \]
where \( t \) is the time in seconds after the flare is launched. We need to determine how long it will take for the flare to return to the water. This occurs when the height \( h(t) = 0 \).

#### Solution:
Set \( h(t) = 0 \):
\[ -15t^2 + 150t = 0 \]

Factor out the common term \( -15t \):
\[ -15t(t - 10) = 0 \]

This gives us two solutions:
\[ -15t = 0 \quad \text{or} \quad t - 10 = 0 \]
\[ t = 0 \quad \text{or} \quad t = 10 \]

The solution \( t = 0 \) corresponds to the initial launch of the flare. The solution \( t = 10 \) corresponds to the time when the flare returns to the water.

Thus, the flare will return to the water after:
\[ \boxed{10} \text{ seconds} \]

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Problem 2: Osprey Diving Towards Water

The height of the osprey above the water is given by the function:
\[ h(t) = 5t^2 - 30t + 45 \]
We need to determine the time it takes for the osprey to reach a height of 20 meters. This means we need to solve for \( t \) when \( h(t) = 20 \).

#### Solution:
Set \( h(t) = 20 \):
\[ 5t^2 - 30t + 45 = 20 \]

Rearrange the equation to standard quadratic form:
\[ 5t^2 - 30t + 45 - 20 = 0 \]
\[ 5t^2 - 30t + 25 = 0 \]

Divide the entire equation by 5 to simplify:
\[ t^2 - 6t + 5 = 0 \]

Factor the quadratic equation:
\[ (t - 1)(t - 5) = 0 \]

This gives us two solutions:
\[ t - 1 = 0 \quad \text{or} \quad t - 5 = 0 \]
\[ t = 1 \quad \text{or} \quad t = 5 \]

Thus, the osprey reaches a height of 20 meters at:
\[ \boxed{1 \text{ second and } 5 \text{ seconds}} \]

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Problem 3: Baseball Popped Upwards

The height of the baseball above the ground is given by the function:
\[ h(t) = -16t^2 + 48t + 3 \]
where \( t \) is the time in seconds. The catcher catches the ball when it is 3 feet above the ground. We need to determine how long the ball was in the air.

#### Solution:
Set \( h(t) = 3 \):
\[ -16t^2 + 48t + 3 = 3 \]

Subtract 3 from both sides:
\[ -16t^2 + 48t + 3 - 3 = 0 \]
\[ -16t^2 + 48t = 0 \]

Factor out the common term \( -16t \):
\[ -16t(t - 3) = 0 \]

This gives us two solutions:
\[ -16t = 0 \quad \text{or} \quad t - 3 = 0 \]
\[ t = 0 \quad \text{or} \quad t = 3 \]

The solution \( t = 0 \) corresponds to the initial launch of the ball. The solution \( t = 3 \) corresponds to the time when the ball is caught by the catcher.

Thus, the ball was in the air for:
\[ \boxed{3} \text{ seconds} \]

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Problem 4: Surf Shop Profit

The profit of Angela's surf shop is modeled by the function:
\[ P(t) = 1125(t-1)^2 - 10,125 \]
where \( t \) is the number of years of operation. We need to determine how long it will take for the shop to start making a profit. This occurs when \( P(t) > 0 \).

#### Solution:
Set \( P(t) = 0 \) to find the break-even points:
\[ 1125(t-1)^2 - 10,125 = 0 \]

Add 10,125 to both sides:
\[ 1125(t-1)^2 = 10,125 \]

Divide both sides by 1125:
\[ (t-1)^2 = 9 \]

Take the square root of both sides:
\[ t-1 = \pm 3 \]

This gives us two solutions:
\[ t-1 = 3 \quad \text{or} \quad t-1 = -3 \]
\[ t = 4 \quad \text{or} \quad t = -2 \]

Since \( t \) represents the number of years of operation, we discard the negative solution \( t = -2 \). The shop starts making a profit when \( t > 4 \).

Thus, it will take:
\[ \boxed{4} \text{ years} \]

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Final Answers:

1. \(\boxed{10}\)
2. \(\boxed{1 \text{ second and } 5 \text{ seconds}}\)
3. \(\boxed{3}\)
4. \(\boxed{4}\)