Question #afdc8

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Question #afdc8

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Answer 1 · 2016-09-15T13:15:34

$u \to Initial upward vlocity of the object dropped = + 9 \frac{m}{s}$ $u->"Initial upward vlocity of the object dropped"=+9m/s$

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

$h \to Vertical displacement of the object = ?$ $h->"Vertical displacement of the object"=?$

By equation of kinematics

$- h = 9 \times 18 - \frac{1}{2} \times 9.8 \times 18^{2}$ $-h=9xx18-1/2xx9.8xx18^2$

$h = 1425.6 m$ $h=1425.6m$

So helicopter was 1425.6m high when the object was dropped

Answer 2 · 2016-09-15T13:40:45

$sf(1430color(white)(x)m)$

Explanation:

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

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

Since g acts vertically this becomes:

$sf(h=ut-1/2"gt"^2)$

$:.$ $sf(h=(9xx18)-1/2(9.81xx18^2)$

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

$sf(h=-1427color(white)(x)m)$

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.

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Question #afdc8

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