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`        how to find velocity at half of maximum height provided velocity is v and range is max`
3 years ago

Vikas TU
10603 Points
```							Maximum Height = v^2/2gHalf of Max. Height = H/2 =>H’ = v^2/4gRange is maximum means thetha of projection = 45 degree.Therfore at half of max. height = > H = u^2/8gu^2 = 8gHu = root(8gH) m/s
```
3 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