Let Δt be the time it takes for the projectile to travel.
The equation Δx = vΔt only applies for one-dimensional kinematics. In 2D kinematics, the v is the x component of velocity. Just because velocity has been halved doesn't mean that the x component of velocity has been halved.
The upward force should be the tension in the cord that holds the pulley up. The downward force is the tensions in the ropes (technically, 2*15.6N since there are two ropes hanging from the pulley). And the net force is 0.
Ah, I see. I just re-read the original question: it's "the tension in the cord that supports the pulley". I was confused slightly by the wording of the question, and didn't have access to the problem's diagram.
The net force on the pulley is 0. Draw a force diagram for the pulley; it is acted...
Technically yes, it would be mg - T = ma. Let "a1" be the acceleration of the first mass and "a2" be the acceleration of the second mass.
We have the equations:
(1) T - m1*g = m1*a1
(2) T - m2*g = m2*a2
Both m1 and m2 undergo the same magnitude of acceleration, but in opposite directions...
To solve these types of problems, I would always recommend starting by drawing force diagrams. This will help you answer such questions as: What upward force is acting on m1? What downward force is acting on m1? What is the net force on m1? What about m2?
You have to make sure your signs are consistent.
Why are you choosing the displacement (x) to be positive but the acceleration (a) to be negative? In this situation, they should be in the same direction (downward).
My apologies -- the original answer was correct.
When the radius is halved, the force increases by a factor of 4. And when both masses are doubled, the force is quadrupled. So the force ultimately increases by a factor of 16.