Question

1. Assume an airplane flew horizontally with a constant velocity and dropped one bale of hay every two seconds to a herd of cattle below. If the air resistance can be ignored, (a) what is the motion of the bales as relative to the ground? (b) what is the motion of the bales as relative to the airplane? (c) As the bales are falling through the air, will their distance of separation before hitting the ground increase, decrease or remain constant?

Answer 1

(a) When bales are released, they have a horizontal velocity and no horizontal acceleration and no vertical velocity and gravitational acceleration vertically downwards  with respect to ground.

So, this motion will be parabolic with respect to ground as there is horizontal velocity and vertical acceleration.

Suppose horizontal speed of plane be v and height be h.

Horizontal distance x = vt => t = x/v

Vertical distance from level of plane y = (1/2)gt² => t = (2y/g)^0.5

From these equations, x/v = (2y/g)^0.5

=> x² = 2v²y/g which is equation of a parabola.

(b) The airplane and bale have same horizontal speed. So relative horizontal speed is 0.

Vertical acceleration is g.

So the bale only has vertical motion wrt this frame. So the motion will be a straight line.

(c) Since, there is no acceleration in horizontal direction, the horizontal distance i.e. x will be constant and equal to

x' = x1 - x2 = vt - v(t-2) = 2v

In vertical direction , y' = y1 -y2 =(1/2) gt² - ( 1/2 ) g (t -2 )² = (1/2)g(4t-4) = 2*g*(t-1), thus this increases with time

Hence net distance between two bales = r' = (x'² + y'² )^0.5 = ( 4v² + 4g²(t-1)² )^0.5 increases with time.

So distance of separation increases with time