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`        A child on a swing  swing is 1 m above the ground at the lowest point and 6m above the ground at highest point the horizontal speed of the child at the lowest point is?`
2 years ago

Arun
24740 Points
```							Loss in P.E = Gain in K.E mgh = 1/2mv^2 g(5) = 1/2v^2 10g = v^2 v = √100 = 10 m/s RegardsArun (askIITians forum expert)
```
2 years ago
Nived Kamal
14 Points
```							At top point potential energy equals 5mg if we consider 1m as the zero potential region.Velocity there is zero.At bottom point kinetic energy is 1\2mv^2.potential energy is zero.Applying conservation of mechanical energy on these points,we get, velocity at bottom equals 10m/s.
```
2 years ago
Siddharth Singh
100 Points
```							Loss in P.E = Gain in K.E mgh = 1/2mv^2 g(5) = 1/2v^2 10g = v^2 v = √100 = 10 m/sThanks from Siddharth Singh
```
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

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