How much farther does a free falling object fall in 10 seconds than in 1.0 seconds?
(divide one distance by the other to get the ratio)
distance fell is proportional to time squared. what is 10^2 and 1^2
thanks!
To calculate the distance a free-falling object falls in a given time, we can make use of a physics formula. According to the laws of motion, the distance an object falls can be determined using the following equation:
d = 0.5 * g * t^2
Where:
d is the distance
g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
t is the time in seconds
For the given question, we need to find the distance fallen in two different time intervals: 10 seconds and 1.0 second.
Let's calculate the distance fallen in 10 seconds:
d1 = 0.5 * g * (10)^2
d1 = 0.5 * 9.8 * 10^2
d1 = 0.5 * 9.8 * 100
d1 = 490 meters (rounded to the nearest meter)
Now, let's calculate the distance fallen in 1.0 second:
d2 = 0.5 * g * (1.0)^2
d2 = 0.5 * 9.8 * 1.0^2
d2 = 0.5 * 9.8 * 1.0
d2 = 4.9 meters
To find the ratio between the two distances, we can divide the distance fallen in 10 seconds (d1) by the distance fallen in 1.0 second (d2):
ratio = d1 / d2
ratio = 490 / 4.9
ratio = 100
Therefore, a free-falling object falls 100 times farther in 10 seconds than it does in 1.0 second.