A small flat cushion of mass m is released from rest at a top corner of a buildi

Question

A small flat cushion of mass m is released from rest at a top corner of a building having height h. A wind blowing along the side of the building exerts a horizontally forward force of constant magnitude B on the cushion as it drops. The air exerts no vertical force. (a) Show that the path of the cushion is a straight line. (b) Does the cushion fall with constant velocity? Explain. (c) If m = 1.20 kg, h = 8.00 m, and B = 240 N, then how far from the building will the cushion hit the level ground?

Explanation / Answer

a)

The position of an object along a straight line can be uniquely identified by its distance from a origin. since the cushion is free falling its path will be a straight line.Even if the wind is acting horizontaly its weight acting downward.so it will have a linear downward motion in a straight line.

b)

free fall is any motion of a body where gravity is the only force acting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it .Cushion falls with constant velocity.

c)

S = 1/2 gt^2 = 8 m the height of the building/fall; g = 9.81 m/sec2 and

time ,t = sqrt(2S/g) = sqrt(2*8/9.8)=1.27s

substituting time in the below equation

X = 1/2 at^2; where F/m = a is the horizontal acceleration due to the wind force F = 2.40 and m = 1.20 kg.

a=2m/s2

X = 1/2 *2*1.272 = 1.61m

which is the distance from the building

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