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?

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

is given by

where y=displacement

u=initial velocity

a=acceleration

t=time

for object B

Subtract 1 and 2 we get

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

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s