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 1.50 s to reach the water.
A. How high above the water was the preserver released?
B. List the knowns in this problem.

Answer 1

Answer:

13.125 m

Explanation:

b) The quantity which are known as

Initial velocity of life preserver, u = 1.40 m/s

time to reach, t = 1.50 s

final velocity of the life preserver, v = 0 m/s

acceleration due to gravity = 9.8 m/s²

a) height of the preserver above water

   using equation of motion

    $h = u t + \dfrac{1}{2}gt^2$

    $h = 1.4\times 1.50 + \dfrac{1}{2}\times 9.8\times 1.5^2$

          h = 13.125 m

Hence, the height from where preserver is thrown is equal to 13.125 m