QuestionCalculate the acceleration due to gravity at the top of a 281-meter building using $g(h)=\frac{3.99 \times 10^{14}}{(6.374 \times 10^{6}+h)^{2}}$.

Studdy Solution
Perform the division to find the acceleration due to gravity at the top of the building.
$$g(281)=\frac{3.99 \times10^{14}}{4.063 \times10^{13}} =9.8123 \, m/sec^{2}$$So, the acceleration due to gravity at the top of the building is approximately $9.8123 \, m/sec^{2}$.