Question

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.

Answer 1

The answer is 9 m.

Step-by-step explanation:

Using the kinematic equation for an object in free fall:

`y = y_o - v_o-(1)/(2)gt^(2)`

In this case:

`v_o = \textrm{Initial velocity} = 1.5[m/s]\\t = \textrm{air time} =  1.2 [s]`
`y_o = 0`

`g = \textrm{gravity} = 9.8 [m/s^(2) ]`

Plugging those values into the previous equation:

`y = 0 - 1.5*1.2-(1)/(2)*9.8*1.2^(2) \\y = -8.85 [m] \approx -9 [m]`

The negative sign is because the reference taken. If I see everything from the rescuer point of view.