An object is thrown straight down from the top of a building 200 feet tall with an initial velocity of 20 ft/sec.
Use the position function s(t) = –16t 2 + v0t + s0.
use it for what? I guess you can start with
s(t) = 200 - 20t - 16t^2
now let 'er rip
To use the position function s(t) = -16t^2 + v0t + s0, we need to understand the variables involved:
- s(t) represents the position of the object at time t.
- t denotes the time elapsed since the object was thrown.
- v0 represents the initial velocity of the object.
- s0 denotes the initial position of the object.
In this scenario:
- The object is being thrown straight down, so its initial velocity, v0, will be negative.
- The initial velocity, v0, is given as -20 ft/sec.
- The initial position, s0, is the height of the building, which is 200 feet. Since the object is being thrown from the top of the building and moving downward, its initial position will also be negative.
Now, we can plug in the values into the equation s(t) = -16t^2 + v0t + s0.
s(t) = -16t^2 - 20t + 200
This equation will give the position of the object at any given time t.