If an object is dropped from a 277-foot-high building, its position (in feet abo

Question

If an object is dropped from a 277-foot-high building, its position (in feet above the ground) is given by d(t) = - 16t^2 + 277, where t is the time in seconds since it was dropped. Compute d'(t). What units are associated with the derivative, and what does it measure? When will the object hit the ground? What is the velocity of the object when it strikes the ground (in mi/hr)? d'(t) = -32t The units associated with the derivative are ft/s and it measures the velocity of the object. The object will hit the ground 4.16 seconds after being dropped. (Round to two decimal places as needed.) The velocity of the object when it strikes the ground is mi/hr. (Round to one decimal place as needed.)

Explanation / Answer

b)

d(t) = 0

-16t2 + 277 = 0

t2 = 277/16

t = 4.16 s

So the object will hit the ground after 4.16 s.

c)

d'(t) = -32t

d'(4.16) = -32*4.16

= -133.12 ft/s

= -90.76 miles/hr

So the obejct will hit ground with a velocity of 90.76 miles/hr

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