a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
Answer:
Same weight so neither is heavier
It is the very last answer
The answer is no. If you are dealing with a conservative force and the object begins and ends at the same potential then the work is zero, regardless of the distance travelled. This can be shown using the work-energy theorem which states that the work done by a force is equal to the change in kinetic energy of the object.
W=KEf−KEi
An example of this would be a mass moving on a frictionless curved track under the force of gravity.
The work done by the force of gravity in moving the objects in both case A and B is the same (=0, since the object begins and ends with zero velocity) but the object travels a much greater distance in case B, even though the force is constant in both cases.