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`        maximum height reached by projectile is 4m. The horizontal range is 12m. velocity of projection?`
2 years ago

## Answers : (1)

Arun
23739 Points
```							Dear Student H  = (R tan Ф ) / 4 4  = (12 tan Ф ) / 4tan Ф = 4/3        sin 2Ф = 2 tan Ф / (1 + tan² Ф)  = 0.96R = u² sin 2Ф / g  u² =  12 * g / 0.96  = 25 g /2 u = 5 √(g/2) RegardsArun (askIITians forum expert)
```
2 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