A ball is thrown straight down from the top of a 453-foot building with an initial velocity of -21 feet per second.

Question

A ball is thrown straight down from the top of a 453-foot building with an initial velocity of -21 feet per second. Use the position function below for free-falling objects.

s(t) = -16t² + v₀t + s₀

What is its velocity after 3 seconds?

V(3)= ft/s

What is its velocity after falling 340 feet?

V= ft/s

Answer 1

h(3) = -16t^2 -21t + 453

= -16(9) -21(3) + 453

= -144 -63 + 453

= -207 + 453

= 246 feet high after 3 seconds

v= h'(t) = -32t -21

v(3) = h'(3) = -32(3) -21 = -96 - 21 = - 117 feet per second velocity after 3 seconds

h(t) = 340 = -16t^2 -21t + 353

-16t^2 -21t + 13 = 0

16t^2 +21t -13 = 0

t = -21/32 + or - (1/32)sqr(21^2 + 64(13)) = -21/32 + or - (1/32)35.7 = 14.7/32 = .46 seconds

v(.46) = h'(.46) = -32(.46) - 21 = -35.7 feet per second velocity after 0.46 seconds