# A 20.0 kg child is on a swing that hangs from 3.00 m -long chains?

A 20.0 kg child is on a swing that hangs from 3.00 m-long chains. What is her speed v(i) at the bottom of the arc if she swings out to a 45.0 degree angle before reversing direction? The answer is in m/s.

So far I came up with 6.45 and 18 which are both incorrect. Please help

**2** answer

If she swings out to 45° then the vertical component of the swing chains will be 3cos45° so the amount she will have risen will be 3 - 3cos45° which is about 0.88m

Now from that you can work out her potential energy which will be mass x height = 20 x 0.88 x g = 172.48J

Assuming no losses that means that her kinetic energy at the bottom of the swing will be the same, so from 172.48 = 1/2mv^2

v^2 = 17.248 so v= 4.15m/s

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A 20.0 kg child is on a swing that hangs from 3.00 m -long chains?

A 20.0 kg child is on a swing that hangs from 3.00 m-long chains. What is her speed v(i) at the bottom of the arc if she swings out to a 45.0 degree angle before reversing direction? The answer is in m/s.

So far I came up with 6.45 and 18 which are both incorrect. Please help