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Two particles of the same mass are projected from the ground with the same speed 10 m/s at an angle 37 and 53 with the horizontal respectively. Find the maximum height attained by their COM (in metres). Take (g=9.8 m/s2) and answer upto two decimal places. 


Solution


Horizontal components of velocity are 
u1x=u1cosθ=10×cos37=8 m/s
u2x=u2cosθ=10×cos53=6 m/s
and
Vertical component of velocity are 
u1y=u1sinθ=10sin37=6 m/s
u2y=u2sinθ=10sin53=8 m/s

Let mass of each particle be m.
Then,
Vertical component of velocity of COM is 
(vycom)=u1ym1+u2ym2m1+m2
=6m+8m2m=7 m/s

Maximum height attained by COM is
Hcom=(vycom)22g=2.5 m 

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