Question

A packet is dropped from helicopter flying at an altitude of 1960 m above the surface of the Earth how much time will it take to reach the ground if the helicopter was stationery.

Answer 1

From conservation of angular momentum we can derive angular velocity as a function of height:

ω(h)=ω0(R+h0R+h)2
ω(h)=ω0(R+h0R+h)2
where ωoωo is the angular velocity of the earth, RR is the radius of the earth, hh is the current height and h0h0 is the initial height.

The horizontal velocity (in the frame of reference of the earth) is

v(h)=(ω0−ω(h))(R+h)=ω0(R+h)−ω0(R+h0)2(R+h)
v(h)=(ω0−ω(h))(R+h)=ω0(R+h)−ω0(R+h0)2(R+h)
A bit more manipulation gives

v(h)=ω0R+h((R+h)2−(R+h0)2)≈ω02R(h−h0)R+h≈2ω0(h−h0)
v(h)=ω0R+h((R+h)2−(R+h0)2)≈ω02R(h−h0)R+h≈2ω0(h−h0)
(The approximations are valid because R>>hR>>h).

In other words, the velocity scales directly with the difference in height. As a function of time, we write

v(t)=2ω0((h0−12gt2)−h0)=−omega0gt2
v(t)=2ω0((h0−12gt2)−h0)=−omega0gt2
Now we integrate:

x=∫T0−ω0gt2dt=−13ω0gT3
x=∫0T−ω0gt2dt=−13ω0gT3
where TT is the time taken for the fall, roughly T=2h0g−−−√T=2h0g
Substituting, we get



x=−13ω0g(2h0g)32=24cm
x=−13ω0g(2h0g)32=24cm
Since you did not tell us what the "correct" answer was, or what the details of your math were, I can't tell whether this solves your problem