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`        An object is projected with velocity V. If the range R of this object is double the maximum height H, then its range is1. V^2/g2. 3v^2/5g3. 4v^2/5g4. V^2/2g`
2 years ago

Vikas TU
9499 Points
```							Range = V^2*sin(2*thetha)/gandH = V^2*(sin(thetha))^2/2ggiven, R = 2HTherefore,V^2*sin(2*thetha)/g = V^2*(sin(thetha))^2/2g=> 2sin(thetha)cos(thetha) = sin(thetha)^2/2 on solving we get sinthetha = 4/root(17)and the Range => 8V^/17g
```
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