the distance s that an object falls is directly proportional to the square of the time t of the fall. if an object falls 16 feet in 1 second, how far will it take an object to fall 64 feet?

Hi Crys;

*the distance s that an object falls is directly proportional to the square of the time t of the fall. if an object falls 16 feet in 1 second, how far*

**long**will it take an object to fall 64 feet?

s=(1/2)gt

^{2}Acceleration due to gravity is 9.8 m/s

^{2}.Our first priority is to convert this into feet to correspond to the question you present.

Acceleration due to gravity is 32 ft/s

^{2}Henceforth, in 1 second, this object fell...

x=(1/2)(32 ft/s

^{2})(1 s)^{2}x=(1/2)(32 ft/s

^{2})(1 s^{2})The unit of s

^{2}is in the denominator and numerator. It cancels...x=(1/2)(32 ft)

The only unit remaining is feet. This is what we want.

x=16 feet

64 ft=(1/2)(32 ft/s

^{2})(t^{2})64 ft=(16 ft/s

^{2})(t^{2})Let's divide both sides by (16 ft/s

^{2})...(64 ft)/(16 ft/s

^{2})=t^{2}The unit of feet is in the numerator and denominator. It cancels.

The unit of s

^{2}is in the denominator of the denominator. It is now in the numerator.(64/16)(s

^{2})=t^{2}4 s

^{2}=t^{2}Square root both sides...

2 s=t

**It will require 2 seconds for it to fall 64 feet.**

## Comments

^{2}, 32 feet/s^{2}. Size is a non-issue. Weight is an issue for maximum velocity only because the greater the weight the greater the ability to fight air resistance.