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Q44PE

Expert-verifiedFound in: Page 83

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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.**

b) 18.396 m

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/s
^{2}. - Final position = 0.

**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.

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