Question

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B falls through a distance Db during a time 2t. If air resistance is negligible, what is the relationship between Da and Db?Da=1/4DbIt cannot be determined from the information given.Da=4DbDa=2DbDa=1/2Db

Answer 1

Da=(1/4)Db

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

`s=ut+(1)/(2)at^2\\\Rightarrow Da=0* t+(1)/(2)* a* t^2\\\Rightarrow Da=(1)/(2)at^2`

When s = Db, t = 2t

`s=ut+(1)/(2)at^2\\\Rightarrow Da=0* t+(1)/(2)* a* (2t)^2\\\Rightarrow Db=(1)/(2)a4t^2`

Dividing the two equations

`(Da)/(Db)=((1)/(2)at^2)/((1)/(2)a4t^2)=(1)/(4)\\\Rightarrow (Da)/(Db)=(1)/(4)\\\Rightarrow Da=(1)/(4)Db`

Hence, Da=(1/4)Db