Question

At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the roof. During their descent to the ground what happens to the distance between the two objects?

Answer 1

Answer:

Explanation:

Given two objects are dropped simultaneously

Object A is 10 m higher than object B therefore

Distance covered by object A is given by

$y_a(t)$ is given by

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

$y_a(t)=0+0.5gt^2--1$

for object B

$y_b(t)=0+0.5gt^2--2$

Subtract 1 and 2 we get

$y_a(t)-y_b(t)=0$

i.e. they will travel equal distance in equal time and distance between them remain 10 m until object B hits the ground