Two particles of the same mass are projected from the ground with the same speed 10m/s at an angle 37∘ and 53∘ with the horizontal respectively. Find out the maximum height attained by their COM. (g=10m/s2)
A
4.9m
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B
5m
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C
2.45m
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D
10m
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Solution
The correct option is C2.45m
Horizontal components of velocity are u1x=u1cosθ=10×cos37∘=8m/s u2x=u2cosθ=10×cos53∘=6m/s
and
Vertical component of velocity are u1y=u1sinθ=10sin37∘=6m/s u2y=u2sinθ=10sin53∘=8m/s
Let mass of each particle be m.
Then,
Vertical component of velocity of COM is (vycom)=u1ym1+u2ym2m1+m2 =6m+8m2m=7m/s
∴ Maximum height attained by COM is Hcom=(vycom)22g=2.45m