Question #afdc8

2 Answers
Sep 15, 2016

#u->"Initial upward vlocity of the object dropped"=+9m/s#

#t->"Time of flight of the object"=18s#
#g->"Acceleration due to gravity"= -9.8ms^-2#

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

By equation of kinematics

#-h=9xx18-1/2xx9.8xx18^2#

#h=1425.6m#

So helicopter was 1425.6m high when the object was dropped

Sep 15, 2016

Answer:

#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.