Question

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

Answer 1

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

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