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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.
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
```
7 years ago

147 Points
```										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
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
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