A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.
It helps because it's being transported blah blah whatever the last person said when you first asked this question
Explanation:
Given that,
The initial velocity of a skater is, u = 5 m/s
She slows to a velocity of 2 m/s over a distance of 20 m.
We can find the acceleration of skater. It is equal to the rate of change of velocity. So, it can be calculated using third equation of motion as follows :

a = acceleration

So, her acceleration is
and she is deaccelerating. Also, her initial velocity is given i.e. 5 m/s.
A light year is the DISTANCE light travels through vacuum in 1 year.
If light is traveling through vacuum, then it's traveling at the speed of light in vacuum. If a student at home at the beginning of the trip is holding the clock, then ...
Traveling 1 light year takes 1 year.
Traveling 2 light years takes 2 years.
Traveling 3 light years takes 3 years.
Traveling 10 light years takes 10 years.
If the light is traveling through some other substance, or if the clock is traveling along with the light, then these numbers all change.
YOU cannot travel at the speed of light. We have to just leave it at that