A helicopter is 400 ft above the ground, and is moving horizontally at 50 ft/s At what rate is the angle (inradians) between the string and the horizontal decreasing when 200ft of string have been let out? (Enter your answer as afraction.)

1rad/s

the rate of change of the angle is - 767/100 rad/s

Step-by-step explanation:

Assuming that the string is released at the origin (0,0) from rest, then the length of the string is:

y = 1/2*g*t²

x = v*t

then

L = √ (x²+y²) = √[ (1/2*g*t²) ² + (v*t) ²]

when L=200 m

L² = (1/2*g) ² * t⁴ + v² * t²

for L = 200 m

doing z=t²

(1/2*g) ² * z² + v² * z - L² = 0

(1/2 * 32.174 ft/s²) * z² + (50 ft/t) ² * z - (200ft) ² = 0

16.087 ft/s² * z² + 2500 ft²/s² * z - 40000 ft² = 0

z = 14.62 s²

t = √z = √ (14.62 s²) = 3.82 s

since

tg θ = (H-y) / x = H / (v*t) - g / (2*v) * t

θ = tan⁻¹ (H / (v*t) - g / (2*v) * t)

since the rate of change equals the derivative with respect to the time

dθ/dt = (-1) / [1 + (g / (2*v) * t) ²] * [ - H / (v*t²) - g / (2*v) ]

dθ/dt = [-k - H / (v*t²) ]/[1 + [H / (v*t) - k*t]²], k = g / (2*v)

since k = g / (2*v) = 32.174 ft/s² / (2*50 ft/s) = 0.32174 s⁻¹, t = 3.82 s

dθ/dt = [-k - H / (v*t²) ]/[1 + [H / (v*t) - k*t]²] = - [0.32174 s⁻¹ + 400 ft / (50 ft/s*3.82 s) ] / [ 1 + (400 ft / (50ft/s*3.82 s) - 0.32174 s⁻¹ * 3.82 s) ²]

dθ/dt = - 7.67 s⁻¹ = - 767/100 rad/s