Consider a tall building located on the Earth\'s equator. As the Earth rotates,
ID: 1349675 • Letter: C
Question
Consider a tall building located on the Earth's equator. As the Earth rotates, a person on the top floor of the building moves faster than someone on the ground with respect to an inertial reference frame because the person on the ground is closer to the Earth's axis. Consequently, if an object is dropped from the top floor to the ground a distance h below, it lands east of the point vertically below where it was dropped.
(a) How far to the east will the object land? Express your answer in terms of h, g, and the angular speed of the Earth. Ignore air resistance and assume the free-fall acceleration is constant over this range of heights.
x =
(b) Evaluate the eastward displacement for h = 54.5 m.
Explanation / Answer
We know that the angular speed is = Linear speed/radius
w = V1/R1 = V2/(R1+h)
So , let us consider that the angular velocity is w
and the top of the buliding is at radius R1 and the ground will be at R1+h
since angular velocity remain same for both therefore
V1/R1 = V2 /(R1+h)
V2 - V1 = (V1/R1)h = wh
Now the time taken by the object to cover the distance h
h= (1/2)gt2
t = (2h/g)1/2
so horizontal distance
= time*horizontal speed
= (2h/g)1/2*(V2 - V1)= wh(2h/g)1/2
(b) Eastward displacement
= wh(2h/g)1/2 = w(2h3/g)1/2 = 181.67w metre
w is not given in the question