A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.50 m/s and observes that it takes 1.2 s to reach the water. How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. Round your answer to the nearest whole number.

In order to calculate the height from which the life preserver was released, we can use the following kinematic equation which describes the motion of an object undergoing constant acceleration:

\[ s = ut + \frac{1}{2}at^2 \]

Where: - \( s \) is the distance traveled (which is the height we are looking for), - \( u \) is the initial velocity, - \( t \) is the time it takes to travel, - \( a \) is the acceleration due to gravity (which we will assume to be \( 9.81 \, m/s^2 \)).

Given from the question: - \( u = 1.50 \, m/s \) (since it is thrown straight down, this is positive), - \( t = 1.2 \, s \), - \( a = 9.81 \, m/s^2 \) (the acceleration due to gravity).

Now we can plug these values into the kinematic equation:

\[ s = (1.50 \, m/s)(1.2 \, s) + \frac{1}{2}(9.81 \, m/s^2)(1.2 \, s)^2 \]

First, calculate the components: \[ (1.50 \, m/s)(1.2 \, s) = 1.80 \, m \] \[ \frac{1}{2}(9.81 \, m/s^2)(1.2 \, s)^2 = \frac{1}{2}(9.81 \times 1.44 \, s^2) \] \[ \frac{1}{2}(9.81 \times 1.44 \, s^2) = \frac{1}{2}(14.1164 \, m^2/s^2) \] \[ \frac{1}{2}(14.1164 \, m^2/s^2) = 7.0582 \, m \]

Now, add both components to find \( s \): \[ s = 1.80 \, m + 7.0582 \, m \] \[ s = 8.8582 \, m \]

Rounding to the nearest whole number: \[ s \approx 9 \, m \]

So, the life preserver was released from approximately 9 meters above the water.

Extra: When studying the motion of objects under gravity, it is important to understand a few basic kinematic equations. These equations are useful when predicting the position, velocity, and acceleration of objects in motion when the acceleration is constant. In this case, we consider the acceleration due to gravity to be constant (approximately 9.81 m/s^2 on the surface of the Earth). We have neglected air resistance due to the downdraft of the helicopter. This is not always the case in real-life scenarios, where air resistance can significantly affect the motion of falling objects, especially if they have a large surface area relative to their mass. However, for dense objects and in situations where air resistance is minimized, it is reasonable to use the simple equations of kinematics, as we did in this example.