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`        A ball loses 15% of its kinetic energy after it bounces back from a concrete slab . The speed with which one must throw to have it. Bounce back to same height ‘h’ is . Here h is equal to 12.4 m . Kindly give explanation to each step you do `
one year ago

Arun
24740 Points
```							Dear Vansh I have attached an image containing solution. Please check and let me know if any difficulty arises.  RegardsArun
```
one year ago
Khimraj
3007 Points
```							Since the ball will bounce back to the same height Ui = Uf,0.85 (Ki + Ui) = Uf0.85 (Ki + Uf) = UfKi = (0.15 * Uf) / 0.85½ mvi^2 = (0.15 * mgh) / 0.85vi = square root of [(0.15 * 2 * m * g * h) / (0.85 * m)]vi = square root of [(0.15 * 2 * 9.8 * h) / ( 0.85)]vi =  square root of [ 3.45 * 12.4] = 6.54
```
one year ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

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