The table shows the height of a ball that was dropped from a 380-foot tower. What was the ball's rate of fall during the first 3 seconds?
Responses
A 140 feet per second
B 100 feet per second
C 60 feet per second
D 180 feet per second
There is no table provided, so the answer cannot be determined.
the distance fallen is 16t^2 = 144 ft
so, the average speed is 144/3 = 48 ft/s
Apparently the table did not describe free fall on earth.
To determine the ball's rate of fall during the first 3 seconds, we need to calculate the change in height during this time period and divide it by the number of seconds.
Given that the ball was dropped from a 380-foot tower, its initial height is 380 feet. After 3 seconds, the ball would have fallen a certain distance based on the acceleration due to gravity.
The formula for the distance fallen, assuming no air resistance, is:
distance = 0.5 * g * t^2
where g is the acceleration due to gravity (approximately 32.2 feet per second squared) and t is the time in seconds.
By substituting the values into the formula, we can determine the distance fallen:
distance = 0.5 * 32.2 * 3^2
distance = 0.5 * 32.2 * 9
distance = 145.35 feet
The change in height during the first 3 seconds is equal to the initial height minus the distance fallen:
change in height = 380 - 145.35
change in height = 234.65 feet
Finally, we can calculate the rate of fall by dividing the change in height by the number of seconds:
rate of fall = change in height / time
rate of fall = 234.65 / 3
rate of fall ≈ 78.22 feet per second
Based on the calculations, the ball's rate of fall during the first 3 seconds is approximately 78.22 feet per second. None of the given options (A, B, C, D) matches this value, so none of them is correct.
To find the ball's rate of fall during the first 3 seconds, we need to consider the equation of motion. The equation for the height of an object in free fall is given by:
h = h₀ + v₀t - (1/2)gt²
where:
- h is the height of the object at a given time "t"
- h₀ is the initial height (380 feet in this case)
- v₀ is the initial velocity (which is 0 for an object dropped)
- g is the acceleration due to gravity (approximately 32 feet per second squared)
We need to find the change in height, Δh, during the first 3 seconds. To do that, we substitute the values into the equation:
Δh = h - h₀ = (v₀t - (1/2)gt²)
Since the initial velocity is 0, the equation simplifies to:
Δh = -(1/2)gt²
Now we plug in the values:
Δh = -(1/2)(32)(3)² = -(1/2)(32)(9) = -144 feet
The negative sign indicates that the ball fell downward. Therefore, the ball's rate of fall during the first 3 seconds is 144 feet.
Among the given options, none match this value. Therefore, none of the options A, B, C, or D is correct.