Help! Hiker stucked!

Speed of wave in cable: $v = \frac{d}{t} = \frac{15\:\text{m}}{0.25\:\text{s}} = 60\:\text{m/s}.$ The wave speed in a string is $v = \sqrt{F_t/\mu}$ , where $F_t$ is the tension in the string and $\mu$ its linear density in kg/m. For the cable we find $\mu = \frac{m}{\ell} = \frac{8\:\text{kg}}{15\:\text{m}} = \tfrac{8}{15}\:\text{kg/m};$ $F_t = \mu\:v^2 = \tfrac{8}{15}\:\text{kg/m}\cdot \left(60\:\text{m/s}\right)^2 = 1920\:\text{N}.$ This tension force is due to the ("apparent") weight of the load on the cable, $F_t = m g'$ . The gravitational acceleration in the frame of the helicopter and its load is therefore $g' = \frac{F_t}{m} = \frac{1920\:\text{N}}{150\:\text{kg}} = 12.8\:\text{m/s}^2.$ This is the sum of $g + a$ , so that $a = g' - g = 12.8 - 9.8 = \boxed{43.0\:\text{m/s}^2}.$

Note : We have ignored the mass of the cable in the second part of the calculation. Points higher up on the cable support the weight of its lower part. This means that the wave speed increases as the wave gets closer to the helicopter...

Speed of the Transverse Wave is (15/.25)=60 meter per second.

Now,

For Transverse Waves, we know,

Velocity = sqrt(T/m); where "m" is the Mass of the wire per Unit Length.

Here, for the Cable "m" is (8/15) kg per meter &

Tension in the cable is

=(70+80)(9.8)+(70+80)(a); where "a" is the acceleration of the Helicopter.

Because, The Helicopter is accelerating Against Gravity.

Now, plugging in the values in the above equation of velocity, we get,

Tension = (m)(Velocity)^2

(150)(9.8+a) = (8/15)(3600)

Therefore, solving this, we get, a = (12.8-9.8) = 3 meter per (second)^2.

So, the acceleration of the Helicopter is 3 m(s^-2).