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 ......?

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#?

1 Answer
A08
Mar 19, 2017

#25m#

Explanation:

Applicable kinematic expressions for free falling stones are
#s=ut+g t^2# .....(1)
#v=u+g t# ......(2)

  1. 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)
  2. From (2), Velocity of green stone as it crosses red stone
    #v_(1g)=0+10xxt_(1g)#
    #=>v_(1g)=10t_(1g)#
  3. 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
#10sqrt(h_g/5)t=7#
#=>t=7/10sqrt(5/h_g)# ......(4)
4. It is given that after #(t+2) s# distance between the stones becomes #27m#
Again using (1) we get
#s_g-s_r(bar (t+2))=(10t_(1g)(t+2)+10 (t+2)^2)-(0xx(t+2)+10 (t+2)^2)=27#
#=>10t_(1g)(t+2)=27#

Using (3) and (4) we get
#10sqrt(h_g/5)(7/10sqrt(5/h_g)+2)=27#
#=>(7+2xx10sqrt(h_g/5))=27#
#=>20sqrt(h_g/5)=20#
#=>h_g=25m#

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