Q44PE - College Physics (Urone)

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 victime with an initial velocity of  and observes that it takes  to reach the water. 

(a) List the known 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.

Step-by-Step Solution

Answer

b) 18.396 m

Step 1: Given data

a)

The known value is as follows:

• Initial velocity u= $1.40 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ .
• Time is taken t=  1.8 s.
• Acceleration ( in downward direction) =9.8 m/s².
• Final position = 0.

Step 2: Height of the life preserver

The height of the life preserver is equal to the distance.

For finding out the height, we have the equation

 $\mathbf{h}\mathbf{=}\mathbf{ut}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{gt}}^{\mathbf{2}}$

Here h is the height/distance, u is the initial velocity, g is the acceleration, and t is the time.

Substituting the values in the above expression, we get:

$\begin{array}{rcl}\mathrm{h}& =& \left(1.4\right)\times \left(1.8\right)+\frac{1}{2}\left(9.8\right)\times {\left(1.8\right)}^2\\ & =& 18.396 \mathrm{m}\end{array}$

 

Hence the height of the lifesaver is around  18.396 m  above the sea.