A man standing on the edge of the terrace of a high rise building throws a stone vertically up with a speed of 20 m/s. Two seconds later an identical stone is thrown vertically downwards with the same speed of 20 m/s. Then :
Text Solution
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(a,b,c,d)
Relative Intial velocities
u r = 20 – (0) = 20 m/s
Relative acceleration
a r = 0
Relative velocity between them after time
v r = u r + a r .t
= 20m/s
= constant
⇒ is correct
⇒ Since they are thrown from same height
⇒ Speed is same after reaching ground
⇒ Same KE when they hit the ground
⇒ is correct
The time taken by the first stone to come to same height from where it was thrown.
= 
∴ Time interval between two stone when both are at A and going downwards = 4 – 2 = 2s.
Since, relative velocity is Constant between them. So time interval between their hitting the ground = 2 s.
⇒ is correct
Option is obvious from conservation of energy.
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