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Mathematics 18 Online
Adriana03:

A ball is thrown straight down from the top of a 449-foot building with an initial velocity of -15 feet per second. Use the position function below for free-falling objects. s(t) = -16t2 + v0t + s0 What is its velocity after 2 seconds? v(2) = -47 ft/s What is its velocity after falling 316 feet? v = ft/s

Mercury:

old question but will respond so this can be closed they give you: s(t) = -16t^2 + v0_t + s0 where s(t) is the position at (t) seconds, v_0 is the initial velocity (given as -15 ft/s), s0 is the initial position (449 ft) for the first Q: What is its velocity after 2 seconds? ---> plug in t = 2 for the second Q: this is a little trickier. it fell 316 feet, so the *change* in position s(t) - s(t0) equals 316 in terms of the equation, it looks like this: s(t) = -16t^2 + v0_t + s0 subtracting s0 from both sides s(t) - s0 = -16t^2 + v0_t so on the left side, let s(t) - s0 = 316, v_0 stays the same, and solve for t

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