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Grade: 12
        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?
10 months ago

Answers : (3)

Arun
14739 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
 
Regards
Arun (askIITians forum expert)
10 months 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.
						
10 months 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
						
6 months ago
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