Question

A ball is dropped off of a tall building and falls for 6 seconds before landing on the ground. Consider how far the ball falls in its first 3 seconds of free fall (from t = 0 s to t = 3 s) compared to how far it falls in its next 3 seconds (from t = 3 s to t = 6 s).

Answer 1

In the last six seconds the ball will fall 3 times the distance it fell in the first 3 seconds

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

Total distance

`s=ut+(1)/(2)at^2\\\Rightarrow s_1=0* t+(1)/(2)* 9.81* 6^2\\\Rightarrow s=176.58\ m`

At t = 3 seconds

`s=ut+(1)/(2)at^2\\\Rightarrow s_1=0* t+(1)/(2)* 9.81* 3^2\\\Rightarrow s_2=44.145\ m`

In the next 3 seconds it will fall

`s_2=176.58-44.145=132.435\ m`

Dividing the equations

`(s_2)/(s_1)=(132.435)/(44.145)\\\Rightarrow s_2=3s_1`

In the last six seconds the ball will fall 3 times the distance it fell in the first 3 seconds