# A red and a green stone falls freely next to each other along vertical paths at a place of the universe where the acceleration due to gravity is exactly #10 m/s^2# and where nothing prevents free fall. The red stone just starts falling when the ......?

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A red and a green stone falls freely next to each other along vertical paths at a place of the universe where the acceleration due to gravity is exactly #10 m/s^2# and where nothing prevents free fall. The red stone just starts falling when the green stone reaches it. The green stone also started from rest but from a higher position. After a while the distance between the two stones is #7 m# , and then after two more seconds elapses the distance between them will increase to #27 m # . How much higher did the green stone start in #m# ?

A red and a green stone falls freely next to each other along vertical paths at a place of the universe where the acceleration due to gravity is exactly

##### 1 Answer

#### Explanation:

Applicable kinematic expressions for free falling stones are

- Let green stone start from rest from a height
#h_g# above red stone. Time taken by green stone to cross red stone#=t_(1g)# We get

#h_g=0xxt_(1g)+1/2xx10xxt_(1g)^2#

#=>t_(1g)=sqrt(h_g/5)# .....(3) - From (2), Velocity of green stone as it crosses red stone

#v_(1g)=0+10xxt_(1g)#

#=>v_(1g)=10t_(1g)# - It is clear that green stone will always be in the lead.

Let after#ts# of green stone crossing the red stone, distance between the two is#7m#

#s_g-s_r(bart)=(10t_(1g)t+10 t^2)-(0xxt+10 t^2)=7#

#=>10t_(1g)t=7#

Using (3), we get

4. It is given that after

Again using (1) we get

Using (3) and (4) we get