Answer:
Part a)

Part b)

Part c)

Part d)

Explanation:
Part a)
While bucket is falling downwards we have force equation of the bucket given as

for uniform cylinder we will have

so we have


now we have




now we have


Part b)
speed of the bucket can be found using kinematics
so we have



Part c)
now in order to find the time of fall we can use another equation



Part d)
as we know that cylinder is at rest and not moving downwards
so here we can use force balance



Answer:

Explanation:
Since <em>light is so fast</em> we can assume no time passes between the lightning strikes and we observe it. We want to know then how far away did the strike occur for the sound to take 16s to reach our ears. Since the definition of velocity tells us that
, we can write 