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

Answer:

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

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

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s_1=0\times t+\dfrac{1}{2}\times 9.81\times 6^2\\\Rightarrow s=176.58\ m$

At t = 3 seconds

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s_1=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s_2=44.145\ m$

In the next 3 seconds it will fall

$s_2=176.58-44.145=132.435\ m$

Dividing the equations

$\dfrac{s_2}{s_1}=\dfrac{132.435}{44.145}\\\Rightarrow s_2=3s_1$

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