Question

In​ physics, it is established that the acceleration due to​ gravity, g (in meters divided by sec squared) in meters/sec²​, at a height h meters above sea level is given by the formula below. ​$g(h) = \frac{3.99 x 10^{14}}{(6.374 x 10^6 + h)^2}$ where 6.374 x 10⁶⁶ is the radius of Earth in meters.
A certain building is 458 meters tall. What is the acceleration due to gravity at the top at the top of the building​? ​g(h) = _______ m/sec².

Answer 1

Answer:

$g(h)=9.81943\ m/s^2$

Explanation:

The given function is

$g(h)=\dfrac{3.99\times 10^{14}}{(6.374\times 10^6+h)^2}$

Now h = the height from the surface of the Earth

Here the building is 458 m tall

$g(458)=\dfrac{3.99\times 10^{14}}{(6.374\times 10^6+458)^2}$

$\Rightarrow g(458)=9.81943\ m/s^2$

So,

$g(h)=9.81943\ m/s^2$