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Q.Two particles are projected simultaneously from two point A and B such that A is projected from tower of height 5m and B is projected at a distance of 10 m from foot of tower towards A.Both are projected with velocity 10m/s at same inclination 60 degree with horizontal.The time after which their seperation becomes minimum is:
A)2.5s
B)1s
C)5s
D)10s

jiten , 7 Years ago
Grade 11
anser 3 Answers
Arun

Last Activity: 7 Years ago

Dear Jiten
 
I am gettin my answer as 1 sec.
 
Please let me know if this is correct then I will share my solution.
 
Regards
Arun (askIITians forum expert)
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jiten

Last Activity: 7 Years ago

Dear Arun,
Yes answer your answer is correct.Please send the complete detailed answer sir....................................
Thank you..................................................................
Arun

Last Activity: 7 Years ago

Dear student
 
velocity of projection v = 10 m/sec
 
@ = 60 degrees
 
horizontal distance, d = 10 m
Vertical distance h = 5 m
 
For A
 
r1 = (v cos@) t i + [v sin@ – ½ g t^2] j
 
For B
 
r2 = [d – v cos@ t] i + [ – h + v sin@ – ½ g t^2] j
 
so
 
r1 – r2 = [2 v cos @ t – d] + h j
 
r = |r1 – r2 | = sqrt [(2 v cos @ t – d)^2 + h^2 ]
 
 
For r (min)  (parallel condition)
 
dr/dt = 0
 
we will get
 
t = d / 2v cos@
 
putting the values
 
t = 10 / (2 * 10 * cos60) = 1 sec
 
Hope it helps
 
Regards
Arun
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