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```
what is the minmum velocity of a projectile such that it touches 4 vertices of a regular hexagon??

```
11 years ago

18 Points
```							For projectile to pass through four points A, B, C, D of a regular Hexagon of side a,
we must throw the projective from A and its fall at D passing through B and C in between.

for this we must have Range to be twive of maximum height,
i.e    2 * (u^2 * sin^2(@)/ (2 * g))  =  (u^2 * sin2@/  g)
@ = angle of throw.
Solving above  we get @ = arctan(2) = 63.43 degrees.
Putting aobe in any of equations of Range or height we have
u =  4.95 * (a^0.5).
a = side of hexagon.

```
11 years 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