 # 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 following1. The maximum height from horizontal surface attained by projectile.Ans. - 35m2. The velocity of projectile at maximum height.Ans.     5 root 3 m/s Badiuddin askIITians.ismu Expert
148 Points
13 years ago

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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