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A regular hexagon of side 10 root 3 m is kept at horizontal surface . A particle is projected with velocity v m/s at an angle b from horizontal surface such that it will just touch the all four corners of regular hexagon.
Then answer the following
1. The maximum height from horizontal surface attained by projectile.
Ans. - 35m
2. The velocity of projectile at maximum height.
Ans. 5 root 3 m/s
Dear anurag
let projectile is projected at a distance a from the corner of the base so its range must be
R = a+10√3 + a
= 2a +10√3
so 2a + 10√3 = V2 sin2b/2g
or g/v2cos2b = sinb/(2a+10√3) ....................1
select first two corner point of hexagon ,if the projectile pass through these point then by symmetry it will also pass through oteher two ppoints
points are ( a-10√3cos60 , 10√3sin60) and (a,20√3sin60)
use the general equation
y =x tanb - gx2/2v2cos2b
put above these two point and put the value of g/v2cos2b = sinb/(2a+10√3)
then u will get 2 equation in a and tanb
solve it and then easly u can get desired result
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