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. 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.

Answer 1

Answer:

$x=\frac{2h(\omega R)^{2}}{g}$

Explanation:

When the object fall from the top floor it moves with parabolic motion.

This particular case is a horizontal motion of a projectile, so the equation that relates the distance and the time in the y-axes is:

$y=y_{0}+v_{0y}-0.5gt^{2}$

Here: y is the final height, it will be 0, because the object touch the ground.

y₀ is the initial height, it will be h and v₀ in y-axis is zero, because it has a horizontal initial velocity.

So we have:

$0=h+0-0.5gt^{2}$

$h=0.5gt^{2}$ (1)

We know that the velocity at the x-axes remains constant, then we have:

$v_{0x}=\frac{x}{t}$

x is the distance from the base of the building

we can solve it for t and put on (1).

$t=\frac{x}{v_{0x}}$

$h=0.5g\left(\frac{x}{v_{0x}}\right)^{2}$

Solving it for x we will have:

$x=\frac{2hv_{0x}^{2}}{g}$

But $v_{0x}=\omega R[\tex]R is the distance from the center of the earth to the top of the building.ω is the angular speed of the earth.Finally:[tex]x=\frac{2h(\omega R)^{2}}{g}$

I hope it helps you!