use to postion function s(t) = -4.9t^2 + vt + s for free-falling objects
to estimate the height of a building, a stone is dropped from the top of the building into a pool of water at ground level. How high is the building of the splash is seen 6.8 seconds after the stone is dropped?
To estimate the height of the building using the position function for free-falling objects, we need to find the value of s(t) when t = 6.8 seconds.
The position function for free-falling objects is given by s(t) = -4.9t^2 + vt + s, where:
- s(t) represents the position at time t
- -4.9t^2 is the term related to the acceleration due to gravity (which is approximately equal to 9.8 m/s^2)
- vt is the term related to the initial velocity (v) multiplied by time (t)
- s is the initial position (in this case, the height of the building)
In this scenario, since the stone is dropped from the top of the building, the initial velocity (v) would be zero since it starts from rest. Therefore, the position function becomes: s(t) = -4.9t^2 + s
To find the height of the building when the splash is seen 6.8 seconds after the stone is dropped, we need to substitute t = 6.8 seconds into the position function and solve for s.
s(6.8) = -4.9(6.8)^2 + s
Now, we need to know the value of s (the initial position) to calculate the height of the building accurately. If we assume that the stone is dropped from the top of the building (which is a reasonable assumption), we can take s = 0 (since the height of the building is measured from the ground level, and s represents the initial position).
Substituting s = 0 in the equation, we get:
s(6.8) = -4.9(6.8)^2 + 0
Now, we can simply calculate the value of s(6.8) to find the estimated height of the building.