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ADNAN MUHAMMED

Grade 12,

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Plz solve the attachment...,........................ .......

Plz solve the attachment...,........................
.......

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1 Answers

Pratik Kumawat
40 Points
5 years ago
Hey Aniruddha Bohra,
Let’s get to the answer by eliminating options. in option 1 while the first ball reaches ground the distance between them would henceforth decrease and that is not the case. 
Option 2 is also not valid as there should be some initial distance between them while ball 2 is dropped.
Now as we know differtiaton of displacement(s) with respect to time(t) gives velocity. Now when the first ball reaches the ground after that ball 2’s veloity continuously increase. differention also gives the tangent on the curve at some point. So if velocity increases then slope also increses and that is the option 4.
This is for 2nd part of the curve. For first part we know from 2nd equation of motion s=ut+1/2at2. For first at the time of release of second ball there is some velocity which it gained let it be u1. for second ball initial velocity is zero.
applying second eqn. of motion 
s1=u1t+1/2(-g)t2
s2=0+1/2(-g)t2 
|s1-s2|=|u1t|. as you can see distance only depends on t and t varies linearly so first part of graph is straight line which increases with time as t is increasing until first ball reaches the ground.

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