I have no idea on how to get started.
If you do not consider air resistance, gravity causes a falling object to descend a distance of (9.8)t2 where t is the time in seconds. If a 16 pound bowling ball is dropped from the tallest building in the world, how many meters does it fall during the 4th and 5th seconds combined.
Thank you
distance = 9.8t^2
when t=5
distance = 9.8(25) = 245 m
when t=4
distance = 9.8(16) = 156.8
so in the time from 4 seconds to 5 seconds it fell
245-156.8 or 88.2 m
since we are ignoring air resistance, the fact that the ball has a weight of 16 lbs is irrelevant.
what about during the 4th second?
"how many meters does it fall during the 4th and 5th seconds combined"
I saw that Steve.
It did not seem to make a lot of sense to find the total distance of the 4th and 5th second distances.
So I assumed there was a typo and that the question followed the more traditional wording of that type.
To find the distance the bowling ball falls during the 4th and 5th seconds combined, we can use the equation:
distance = (9.8) * t^2
First, let's find the distance for the 4th second. Substitute t = 4 into the equation:
distance during 4th second = (9.8) * (4)^2
distance during 4th second = 9.8 * 16
distance during 4th second = 156.8
So, during the 4th second, the bowling ball falls a distance of 156.8 meters.
Now, let's find the distance for the 5th second. Substitute t = 5 into the equation:
distance during 5th second = (9.8) * (5)^2
distance during 5th second = 9.8 * 25
distance during 5th second = 245
So, during the 5th second, the bowling ball falls a distance of 245 meters.
To find the total distance it falls during the 4th and 5th seconds combined, we add the distances together:
total distance = distance during 4th second + distance during 5th second
total distance = 156.8 + 245
total distance = 401.8
Therefore, the bowling ball falls a total distance of 401.8 meters during the 4th and 5th seconds combined.