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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?

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?

Grade:12

3 Answers

Arun
25750 Points
6 years ago
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)
Nived Kamal
14 Points
6 years ago
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.
Siddharth Singh
100 Points
6 years ago
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

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