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 1.40 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.

a) 0.90 m
b) 1.90 m
c) 2.50 m
d) 4.60 m

Answer 1

To find the height above the water where the preserver was released, we can use the kinematic equation for vertical motion. Plugging in the given values, we find that the preserver was released 18.4 m above the water.

Step-by-step explanation:

To solve this problem, we can use the kinematic equation for vertical motion: y = y0 + v0t + (1/2)gt2. Since the preserver is thrown straight down, the initial velocity (v0) is 1.40 m/s downwards, the time (t) is 1.8 seconds, and the acceleration (g) is -9.8 m/s2 (because it's directed downwards). The unknown variable we're solving for is the initial position (y0) above the water. Plugging in the given values, we have:

y = y0 + v0t + (1/2)gt2

y = y0 + (-1.40 m/s)(1.8 s) + (1/2)(-9.8 m/s2)(1.8 s)2

Simplifying this equation will give us the value of y0, which represents the height above the water where the preserver was released. After calculating, we find that y0 is -18.4 m. However, since height can't be negative, we take the absolute value of y0, which is 18.4 m.