Suppose the free-fall acceleration at some location on earth was exactly 9.8000 m/s2 .what would it be at the top of a 1300 m -tall tower at this location?

Approximately 9.796 m/s^2

The equation for the force that causes the acceleration can be written as.  

Fg = G * M * m ÷ d^2  

where

d is the distance from the center of the earth to the falling object. This distance is equal to the sum of the radius of the earth and the height of the object.  

a = Fg/m  

a = G * M ÷ d^2  

G = 6.67 * 10^-11  

M = 5.98 * 10^24  

G * M = 6.67 * 10^-11 * 5.98 * 10^24 = 3.98866 * 10^14  

9.8 = 3.98866 * 10^14 ÷ d^2  

d^2 = 3.98866 * 10^14 ÷ 9.8  

d = √(3.98866 * 10^14 ÷ 9.8)  

This is approximately 6.38 * 10^6 meters.  

For the top of the building, d = 1300 + √(3.98866 * 10^14 ÷ 9.8)  

a = 3.98866 * 10^14 ÷ [1300 + √(3.98866 * 10^14 ÷ 9.8)]^2  

Which is approximately 9.796 m/s^2.

answer; (d) braking faster then cars;