A helicopter hovering above a forest fire dumps a large bucket of water. How far does the water fall during the third second?
Subtract the distance it fall in 2 seconds from the distance it falls in 3 seconds.
(g/2)*(3^2 - 2^2) = 4.9*5 = 24.5 meters
Well, it depends on how graceful the bucket is feeling that day. But if we assume it's just an average bucket, it might fall somewhere around... let me think... ah, yes, approximately 3 seconds worth of falling distance. Though I must admit, those buckets can be a bit unpredictable sometimes. Just keep an eye out for any water droplets trying to do fancy acrobatics on the way down!
To determine the distance the water falls during the third second, we need to know the acceleration due to gravity and the initial velocity of the water when it is dropped from the helicopter. Assuming negligible air resistance, the acceleration due to gravity can be approximated as 9.8 m/s².
During free fall, an object's velocity increases by 9.8 m/s each second. Therefore, the initial velocity of the water for the third second can be calculated by multiplying the acceleration due to gravity by the time (3 seconds).
Initial velocity of water = acceleration due to gravity × time
= 9.8 m/s² × 3 s
= 29.4 m/s
Now, we can calculate the distance the water falls during the third second by using the formula for distance traveled during free fall:
Distance = Initial velocity × time + (1/2) × acceleration × time²
= (29.4 m/s) × (3 s) + (1/2) × (9.8 m/s²) × (3 s)²
= 88.2 m + (1/2) × 9.8 m/s² × 9 s²
= 88.2 m + 44.1 m
= 132.3 m
Therefore, the water falls approximately 132.3 meters during the third second.
To determine how far the water falls during the third second, we need to calculate the distance traveled by an object in free-fall motion. The formula to calculate the distance fallen is given by:
d = (1/2) * g * t^2
Where:
d is the distance fallen
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time in seconds
In this case, we can assume the helicopter hovers at a constant height, so the time for the water to fall will be the same as the time for the object in free-fall motion.
Thus, we can substitute t = 3 into the formula to calculate the distance fallen during the third second.
d = (1/2) * 9.8 * (3^2)
= (1/2) * 9.8 * 9
= 4.9 * 9
= 44.1 meters
Therefore, the water would fall approximately 44.1 meters during the third second.