Answer:
C. At the instant the ball reaches its highest point.
Explanation:
When a body is thrown up, it tends to come down due to the influence of gravitational force acting on the body. The body will be momentarily at rest at its maximum point before falling. At this maximum point, the velocity of the body is zero and since force acting on a body is product of the mass and its acceleration, the force acting on the body at that point will be "zero"
Remember, F = ma = m(v/t)
Since v = 0 at maximum height
F = m(0/t)
F = 0N
This shows that the force acting on the body is zero at the maximum height.
Efficiency = (Wanted) energy out ÷ energy in × 100
Energy in = 400J
Wanted Energy out = 240J
Energy cannot be used up, only transferred, so the remaining energy is most likely to be transferred into unwanted energy (loss of energy) such as heat energy.
Efficiency = 240 ÷ 400 × 100
Efficiency = 0.6 × 100
Efficiency = 60%
The initial momentum of the system can be expressed as,

The final momentum of the system can be given as,

According to conservation of momentum,

Plug in the known expressions,

Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.
Plug in the known values,

Thus, the final velocity of second mass is 2.99 m/s.
Explanation:
An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.
Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.