SOLUTION: Hi there! I really appreciate the help.
Q: The height h (in feet) of an object falling from a tall building is given by the function h(t)=144-16t^2, where t is the time elapsed
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Q: The height h (in feet) of an object falling from a tall building is given by the function h(t)=144-16t^2, where t is the time elapsed
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Question 1166088: Hi there! I really appreciate the help.
Q: The height h (in feet) of an object falling from a tall building is given by the function h(t)=144-16t^2, where t is the time elapsed in seconds.
a. After how many seconds does the object strike the ground?
b. What is the average velocity of the object from t=0 until it hits the ground?
c. Find the instantaneous velocity of the object after 1 second.
c. Find the instantaneous velocity of the object after 2 seconds.
d. Write an expression for the velocity of the object at a general time a.
v(a)=
e. What is the velocity of the object at the instant it strikes the ground? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! Strikes the ground when h(t)=0
this means 144-16t^2=0
16t^2=144
t^2=9
t=3 seconds
average velocity during that time is (144-0)/(3-0)=48 mph={f(a)-f(b)/b-a)}
v=(1/2)(a(t+h)^2)-(1/2)at^2 divided by h
=(1/2) a (t^2+2th+h^2)-at^2 or 1/2 {at^2+2th+h^2-at^2) divided by h
=(1/2) a (*2th+h^2) divided by h
=(1/2)(2at+ah)
=at+(ah/2)
when h goes to 0 the instantaneous velocity is= at
so at 1 second it is 32 ft/second (32 ft/sec^2*1 sec)
at 2 seconds it is 64 ft/sec
general it is 32 t seconds.
when it strikes the ground v=sqrt(2ax)=sqrt(2*32*144)=96 ft/sec
