A hot air balloon has just lifted off and is rising at the constant rate of 1.8 m/s. Suddenly, one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 11.2 m/s. If the passenger is 2.5 m above her friend when the camera is tossed, how high is she when the camera reaches her? (m)

To solve this problem, we can use the following kinematic equation:

y = y0 + v0y * t - (1/2) * g * t^2

Where:
- y is the final height of the passenger when the camera reaches her,
- y0 is the initial height of the passenger (2.5 m),
- v0y is the initial vertical velocity of the camera (11.2 m/s),
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time it takes for the camera to reach the passenger.

First, let's find the time it takes for the camera to reach the passenger. Since the balloon is rising at a constant rate of 1.8 m/s, the passenger is also moving upward during this time. Therefore, we can assume that both the passenger and the camera have the same time of flight.

Using the equation:
y = y0 + v0y * t - (1/2) * g * t^2

Substituting the given values:
2.5 m = 0 + (11.2 m/s) * t - (1/2) * (9.8 m/s^2) * t^2

Rearranging the equation to solve for t:
0.5 * (9.8 m/s^2) * t^2 - (11.2 m/s) * t + 2.5 m = 0

Solving this quadratic equation will give us two possible values for t. We can use the positive value since time cannot be negative.

Once we have the value of t, we can calculate the final height of the passenger (y) using the same kinematic equation:

y = y0 + v0y * t - (1/2) * g * t^2

Substituting the known values:
y = 2.5 m + (0 m/s) * t - (1/2) * (9.8 m/s^2) * t^2

Simplifying the equation gives us the final height of the passenger when the camera reaches her. Simply plug in the value of t to get the final answer.