A helicopter is traveling with a velocity of 12 m/s directly upward. Directly below the helicopter is a very large and very soft pillow. As it turns out, this is a good thing, because the helicopter is lifting a large man. When the man is 20 m above the pillow, he lets go of the rope.

Question

a. What is the speed of the man just before he lands on the pillow?
b. How long is he in the air after he lets go?
c. What is the greatest height reached by the man above the ground? (HINT: this should be greater than 20 m. Why?)
d. What is the distance between the helicopter and the man three seconds after he lets go of the rope?

Answer 1

There are two motions of the man:
1. upward motion with the initial velocity of 12 m/s
h=vₒ²/2g =144 /2•9.8 =7.34 m
t=vₒ/g=12/9.8=1.22 s.
H=20+7.34 =27.34 m
2. Dowmward motion (free fall)
t=sqrt(2H/g)=sqrt(2•27.34/9.8) =2.36 s.

(a) mgH=mv²/2

v=sqrt(2gH) = sqrt(2•9.8•27.34) = 23.15 m/s.

(b) total t = 1.22+2.36=3.58 s

(c) = H = 27.34 m.
(d) the helicopter ia at height
H1= h+v•t= 20+12•3 = 56 m.
The man:
he is falling during
3-1.22 =1.78 s
and covered distance
h1=gt²/2 = 9.8•1.78²/2 =15.53 m
Since he started from the height H=27.34 m, he is at the height
27.34-15.53 =11.81 m
Δ h =56 – 11.81 = 44.19 m