A ball is thrown straight down from the top of a 489-foot building with an initial velocity of -22 feet per second. Use the position function below for free-falling objects.

Question

s(t) = -16t² + v₀t + s₀

What is its velocity after 2 seconds?

What is its velocity after falling 344 feet?

Answer 1

-16t^2 - 22t + 489 = h(t)

in 2 seconds

-16(2)^2 - 22(2) + 489 = h(2) = -64 -44 + 489 = 381 feet

take the derivative of h(t)

h'(t) = -32t -22

h'(2) = -32(2)-22 = -64-22 = -86 feet per second = velocity after 2 seconds

the -32 is the familiar acceleration due to gravity of 32 feet per second per second = -32 ft/sec^2

velocity is decreasing by -32 ft per second every second. after 2 seconds that's -64 ft/ sec. add that to the initial velocity of -22 to get -86 ft/sec = v(2)

velocity is the derivative of the distance

acceleration (or deceleration) is the derivative of velocity

after falling 344 feet, the time is the solution of:

-16t^2 -22t + 489 = 344

solve for t, then plug that value into the velocity formula: -32t-22 = v(t)

-16t^2 -22t + 145 = 0

16t^2 + 22t -145 = 0

8t^2 + 11t - 72.5 = 0

use the quadratic formula to solve for t

t = -11/16 + or - (1/16)sqr(121+4(8)(72.5) = about 3.8

simplify and plug into v(t) = -32t-22 = v(3.8) -32(3.8) -22 = 143.6 ft/sec