# Question afdc8

Sep 15, 2016

$u \to \text{Initial upward vlocity of the object dropped} = + 9 \frac{m}{s}$

$t \to \text{Time of flight of the object} = 18 s$
$g \to \text{Acceleration due to gravity} = - 9.8 m {s}^{-} 2$

h->"Vertical displacement of the object"=?

By equation of kinematics

$- h = 9 \times 18 - \frac{1}{2} \times 9.8 \times {18}^{2}$

$h = 1425.6 m$

So helicopter was 1425.6m high when the object was dropped

Sep 15, 2016

$\textsf{1430 \textcolor{w h i t e}{x} m}$

#### Explanation:

I will use the convention that "up" is +'ve.

sf(s=ut+1/2"at^2)

Since g acts vertically this becomes:

$\textsf{h = u t - \frac{1}{2} {\text{gt}}^{2}}$

$\therefore$sf(h=(9xx18)-1/2(9.81xx18^2)

sf(h=162-1589.22color(white)(x)m#

$\textsf{h = - 1427 \textcolor{w h i t e}{x} m}$

The minus sign signifies that the ground is 1427 m below the helicopter i.e the helicopter is 1427 m above the ground when the weight is released.