When Li Ming is at point A, the height of kite B is 40 meters above the ground, and the kite line forms an angle of 45 degrees with the horizontal plane, so the horizontal distance between Li Ming and kite B is 40 meters, and the distance between Li Ming and kite B, that is, the line length is 40* 2 meters.

After Li Ming retreats 10 m, pay off at a speed of 0.5 m per second. Find out how long it takes to pay off the line. The kite is just above the original position B, and the kite is at an angle of 60 degrees with the horizontal ground. That is, find the length of the required pay-off.

Suppose that the point where kite B is perpendicular to the ground is point C.

According to the above knowledge

& ltBAC=45 degrees, then AC = BC = 40m, AB=40√2.

DC=DC+AC= 10+40=50

& ltEDC=60 degrees, so

DE=2DC= 100

X seconds later, the kite is just above the original position B at E.

So 0.5x=DE-AB.

That is 0.5x= 100-40√2.

x=200-80√2

x=200-80* 1.4 14

X=97 seconds