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 1.46 m/s and observes that is takes 1.8 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.

Answer 1

18.52 m

Step-by-step explanation:

When the life preserver is dropped from the helicopter, the only force acting on the object is the gravitational force. This modifies the equations of motion. Thus, the working equation is written below:

h = vt + 0.5gt²

where

v is the initial velocity

g is the acceleration due to gravity equal to 9.81 m/s²

h is the height of the fall

h = (1.46 m/s)(1.8 s) + 0.5(9.81 m/s²)
`(1.8 s)^(2)`

h = 18.52 m

**This is the same as the expert answer but they forgot to square the time by two**

Answer 2

When the life preserver is dropped from the helicopter, the only force acting on the object is the gravitational force. This modifies the equations of motion. Thus, the working equation is written below:

h = vt + 0.5gt²
where
v is the initial velocity
g is the acceleration due to gravity equal to 9.81 m/s²
h is the height of the fall

h = (1.46 m/s)(1.8 s) + 0.5(9.81 m/s²)(1.8 s)
h = 11.457 m