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

koraby2bc

Answered question

2022-09-16

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,

Answer & Explanation

skarvama

skarvama

Beginner2022-09-17Added 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 )

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?