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`        A projectile touches 4 verices of a regular hexagon of side length 10√3. Find its time of flight.`
6 months ago

Arun
23389 Points
```							 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√3so 2a + 10√3 = V2 sin2b/2g   or  g/v2cos2b = sinb/(2a+10√3)  ....................1select 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 equationy =x tanb - gx2/2v2cos2bput 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
```
6 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions