*^He@tHeR N
New member
Below is the problem and the answer i got from it... My instuctor told me their was one small error in it.. Can you help me to figure out the right answer?
"Problem:
An object dropped off the top of a tall building falls vertically with constant acceleration. If "s" is the distance
of the object above the ground (in feet) "t" seconds after its release,
then "s" and "t" are related by an equation of the form
s = a + bt^2
where a and b are constants.
So with the above information given, if the object is 240 feet above the ground after 1 second and 192 feet above the ground after 2 seconds.
Find:
(A) Find the constants a and b.
(B) How high is the building?
(C) How long does the object fall?
Solution:
(1.)
S=a+bt^2
case 1: 240 = a+b(1) = a+b.
case 2: 192 = a+b(2)^2 = a+4b
So we have…
a+b = 240 equ 1.
And…
a+2b =(a+ b)+ b = 192
240 +b = 192
Therefore b = - 48.
So,
a = 240 - b = 288 from equ 1.
There for a=288
so, s = 288 -48(t)^2.
(2.) at t= 0,the object is at the top of the building.
so height s = 288 - 48(0) = 288 ft high.
(3.) For time put s= 0 in s = 288- 48( t)^2 and solve for t
So the object falls for t= + or - 4 seconds.
All help is appreciated! THANK YOU SOOO MUCH!!
"Problem:
An object dropped off the top of a tall building falls vertically with constant acceleration. If "s" is the distance
of the object above the ground (in feet) "t" seconds after its release,
then "s" and "t" are related by an equation of the form
s = a + bt^2
where a and b are constants.
So with the above information given, if the object is 240 feet above the ground after 1 second and 192 feet above the ground after 2 seconds.
Find:
(A) Find the constants a and b.
(B) How high is the building?
(C) How long does the object fall?
Solution:
(1.)
S=a+bt^2
case 1: 240 = a+b(1) = a+b.
case 2: 192 = a+b(2)^2 = a+4b
So we have…
a+b = 240 equ 1.
And…
a+2b =(a+ b)+ b = 192
240 +b = 192
Therefore b = - 48.
So,
a = 240 - b = 288 from equ 1.
There for a=288
so, s = 288 -48(t)^2.
(2.) at t= 0,the object is at the top of the building.
so height s = 288 - 48(0) = 288 ft high.
(3.) For time put s= 0 in s = 288- 48( t)^2 and solve for t
So the object falls for t= + or - 4 seconds.
All help is appreciated! THANK YOU SOOO MUCH!!