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 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) 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.

Solution:

We will treat the upward direction as positive, and the downward direction as negative.

Part A

The known values are: a=-9.80 \text{m/s}^2; v_0=-1.40\:\text{m/s}; \Delta t=1.8\:\text{s}; and y_f=0\:\text{m}

Part B

We are looking for the initial position, y_0. We are going to use the formula

\Delta y=v_{0y}t+\frac{1}{2}at^2 
\\
\text{or}
\\
y_f-y_0=v_{0y}t+\frac{1}{2}at^2

Solving for y_0 in terms of the other variables:

y_0=y_f-v_{0y}t-\frac{1}{2}at^2

Substituting the given values:

\begin{align*}
y_0 & =y_f-v_{0y}t-\frac{1}{2}at^2 \\
y_0& =0-\left(-1.4\:\text{m/s}\right)\left(1.8\:\text{s}\right)-\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(1.8\:\text{s}\right)^2 \\
y_0&= 0-\left(-1.4\:\text{m/s}\right)\left(1.8\:\text{s}\right)-\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(1.8\:\text{s}\right)^2 \\ 
y_0& = 0+2.52\:\text{m}+15.876\:\text{m} \\
y_0& =18.396\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}