A stone is dropped from a height of 4.5m.What will be the distance travelled by it during the last second of its fall?
you know that s(t) = 4.9t^2
so, it takes x=sqrt(4.5/4.9) seconds to hit.
You want s(x) - s(x-1)
good answer.
To find the distance traveled by the stone during the last second of its fall, we need to calculate the average velocity of the stone for that last second and then multiply it by the time.
The average velocity can be calculated using the formula:
average velocity = (final velocity + initial velocity) / 2
For a stone dropped from rest, its initial velocity is 0, and we can use the formula for final velocity:
final velocity = initial velocity + acceleration * time
Since the stone is falling under the influence of gravity, we can assume it experiences a constant acceleration equal to the acceleration due to gravity (9.8 m/s^2). Therefore, the final velocity formula simplifies to:
final velocity = 0 + 9.8 * time
To find the time, we know that the stone is dropped from a height of 4.5 m. Using the equation for free fall:
distance = initial velocity * time + (1/2) * acceleration * time^2
Plugging in the known values, we have:
4.5 = 0 * time + (1/2) * 9.8 * time^2
Rearranging this equation, we get:
4.9 * time^2 = 4.5
Simplifying, we find:
time^2 = 4.5 / 4.9
time ≈ 0.96 seconds
Now, we can calculate the final velocity:
final velocity = 0 + 9.8 * 0.96
final velocity ≈ 9.408 m/s
Using the average velocity formula, we can find the average velocity during the last second:
average velocity = (0 + 9.408) / 2
average velocity ≈ 4.704 m/s
Finally, we multiply the average velocity by the time (1 second) to find the distance traveled:
distance = average velocity * time
distance ≈ 4.704 * 1
distance ≈ 4.704 meters
Therefore, the stone will travel approximately 4.704 meters during the last second of its fall.