Question

A kite 100 ft above the ground is moving horizontally at a speed of 8ft/s. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?

Answer 1

the height of the kite h = 100 ft
the horizontal distance of the kite = x = vt where v = 9 ft/s, t is time
h, x and the string (length L) for a right triangle, the angle between the string and the horizontal A satisfies tan A = h/x
sec²A dA/dt = -h/x² dx/dt
we know dx/dt = v
so
dA/dt = -hvcos²A/x² = -hv/(x/cosA)² = -hv/L²
when L = 200 ft,
dA/dt = -100*9/200² = -0.0225 = -9/400 rad/s
negative sign means decreasing.