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`        a ball is dropped onto a floor fom a height of 10m. if 20% of its initial energy is lost , then the height of bounce is (1)2m (2)4m (3)8m (4)6.4m`
3 years ago

Vikas TU
10082 Points
```							mgh = 0.5mv^2v = root(2gh) => root(2*g*10) = > 10root(2) m/sK.E = 0.5m*20080% of KE = >  0.5m*200*80/100 => 80m0.5mv^2 = 80mv^2 = 160From third law of eqn.,0^2 = 160 – 2gs20s = 160s = 8 m
```
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