Derivatives hgdusyge

Question

Consider the position function for free-falling objects.

s(t) = −16t² + v₀t + s₀

A ball is thrown straight down from the top of a 449-foot building with an initial velocity of −26 feet

 per second. Give the corresponding position function s(t).

s(t) = 

−16t2−26t+449

 

Find s'(t).

s'(t) = 

−32t−26

 

Give the corresponding velocity function v(t).

v(t) = 

−32t−26

 

What is its velocity after 4 seconds?

v(4) =  ? ft/s

How many seconds does it take the ball to fall 116 feet?

t =  ? s

What is its velocity after falling 116 feet?

v = ? ft/s

Answer 1

v(4)=-32(4)-26=-128-26=-156 ft/sec in 4 seconds

h(t)=-116=-32t^2-26t

16t^2+13t-58=0

t=-13/32+(1/32)sqr(169+64(58))

= about 1.541 seconds to fall 116 feet

v(1.541) = -32(1.541)-26 = -75.298 ft/sec when it fell 116 feet