A rock is thrown upward from a bridge that is 73 feet above a road. The rock reaches its maximum height above the road 0.98 seconds after it is thrown and contacts the road 3.21 seconds after it was thrown. Write a function ff that determines the rock's height above the road (in feet) in terms of the number of seconds tt since the rock was thrown. f(t)=f(t)= Hint: the function ff can be written in the form f(t)=c*(t−t_1)(t−t_2) f(t)=c*(t-t_1)(t-t_2) for fixed numbers cc, t_1t_1, and t_2t_2.

koraby2bc 2022-09-16 Answered
A rock is thrown upward from a bridge that is 73 feet above a road. The rock reaches its maximum height above the road 0.98 seconds after it is thrown and contacts the road 3.21 seconds after it was thrown.
Write a function ff that determines the rock's height above the road (in feet) in terms of the number of seconds tt since the rock was thrown.
f ( t ) = f ( t ) =
Hint: the function ff can be written in the form f ( t ) = c ( t t 1 ) ( t t 2 )   f ( t ) = c ( t t 1 ) ( t t 2 ) for fixed numbers cc, t1t1,
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Answers (1)

skarvama
Answered 2022-09-17 Author has 5 answers
Let the function f(t) be
f ( t ) = c ( t t 1 ) ( t t 2 )
At t = 3.21 s e c, rock's height = 0 feet
so, t = 3.21 sec is one of the zeroes of the function f(t)
Therefore, f ( t ) = c ( t t 1 ) ( t 3.21 )
At t= 0.98 seconds, rock is at the maximum height i.e. it represents the vertex of the quadratic function f(t).
Since the zeroes are equidistant from the vertex,
( 3.21 0.98 ) = ( 0.98 t 1 )
2.23 = ( 0.98 t 1 )
t 1 = 1.25
So, f ( t ) = c ( t + 1.25 ) ( t 3.21 )
At t = 0 seconds, f(t) = 73
73 = c ( 1.25 ) ( 3.21 )
c = 18.1931
so, f ( t ) = 18.1931 ( t + 1.25 ) ( t 3.21 )

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