Question

consider the position function for free - falling objects. s(t)=-16t^2 + v_0t + s_0. a ball is thrown straight down from the top of a 445 - foot building with an initial velocity of - 15 feet per second. give the corresponding position function s(t). s(t)=x. find s(t). s(t)=x. give the corresponding velocity function v(t). v(t)=. what is its velocity (in ft/s) after 3 seconds? v(3)= ft/s. how many seconds does it take the ball to fall 316 feet? t = s. what is its velocity (in ft/s) after falling 316 feet? v = ft/s.

Explanation:

Step1: Position function with initial values

$s(t) = -16t^2 -15t + 445$

Step2: Derivative of position function

$s'(t) = -32t -15$

Step3: Velocity function is derivative

$v(t) = -32t -15$

Step4: Velocity at t=3

$v(3) = -32(3) -15 = -111$

Step5: Time to fall 316 ft: solve s(t)=445-316=129

$16t^2 +15t -316=0$, solution $t=4$

Step6: Velocity at t=4

$v(4) = -32(4) -15 = -143$

Answer:

$s(t) = -16t^2 -15t + 445$
$s'(t) = -32t -15$
$v(t) = -32t -15$
$v(3) = -111$ ft/s
$t = 4$ s
$v = -143$ ft/s