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?

a. dA = ¼ dB.
b. dA = ½ dB.
c. dA = 2 dB.
d. dA = 4 dB.

Answer 1

a. Da = ¼ Db

Step-by-step explanation:

Let us write out equations of motion for both objects:

For object A:

`D_a = ut + (1)/(2)gt^2`

Since initial velocity, u =, is zero:

`D_a = (1)/(2)gt^2` ---------------------------------------- (1)

For object B:

`D_b = ut + (1)/(2)g(2t)^2\\\\\\D_b = ut + 2gt^2`

Since initial velocity, u =, is zero:

`D_b = 2gt^2` ---------------------------------------- (2)

Note: g = acceleration due to gravity

To get the relationship between Da and Db, we make g subject of formula in (1) and (2) and then equate:

From (1):

`g = (2D_a)/(t^2)`

From (2):

`g = (D_b)/(2t^2)`

Equating:

=>
`(2D_a)/(t^2) = (D_b)/(2t^2) \\\\\\D_a = (D_b)/(4) \\\\\\D_a = (1)/(4) D_b`