I think the problem would be slightly simpler to tackle if you were to attach a coordinate system to the problem. For example, take north as the positive direction and the south as negative.
Now the final direction of motion can be determined by the sign attached to the velocity. For instance...
There is a systematic way to deal with that kind of a problem.
First: Define a coordinate system. Choose a convenient x-axis and y-axis, such that you'd have to take the minmum number of components. And the ones that you take, can be easily found out using some trig.
Second: The component...
That isn't the right thing to say. Work is a scalar quantity given by the dot product(scalar product of two vectors) \vec{W}= \vec{F} . \vec{s} . So it depends on two parameters(viz. the component of the force responsible for the work and the displacement in that direction) and not just one...
Gain in kinetic energy during the course of the motion of the disc, from the instant it was fired till the point when it leaves the barrel; is the sum of the potential energy of the compressed spring and the work done by the barrel(which is negative) If v is required, use:
\frac{1}{2}...
If l is the length of the rope then, the height will be h = l(1 - \cos{\theta} )
The velocity of the boy is then simply \sqrt{2gh} at the equilibrium position. At equilibrium position the velocity of a pendulum is maximum, so that's what you're looking for.
You get the equation by using...
Over the entire trip. i.e up and then back down, it is zero. As I said any work, be it by you or by gravity is dependent only on its initial and final positions, which are same in both cases, so the work is zero.
Work done (in a gravitational field, such as that of the earth) depends only on the initial and final positions, and its path independent.
So if the initial and final positions as in this case are the same, the work will be the same. You don't need to calculate anything.
This question requires you to use 2 concepts, i) that of the Newton' s laws of motion and ii)the conservation of energy.
Firstly, if you've drawn a Free Body Diagram(FBD) correctly then you should get the tension T upwards at the lowermost point in the trajectory(of the the rock), and downwards...