SOLUTION: Thank you for taking the time help with these questions. The path of a falling object is given by the function where represents the initial velocity in ft/sec and represents

Question 135345This question is from textbook College Algebra
: Thank you for taking the time help with these questions.
The path of a falling object is given by the function where represents the initial velocity in ft/sec and represents the initial height in feet. Also, s represents the height in feet of the object at any time, t, which is measured in seconds.
If a rock is thrown upward with an initial velocity of 40 feet per second from the top of a 30-foot building, write the height (s) equation using this information, then determine the height after two (2) seconds, how many seconds to reach maximum height and what is maximum height?

This question is from textbook College Algebra

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The path of a falling object is given by the function where represents the initial velocity in ft/sec and represents the initial height in feet. Also, s represents the height in feet of the object at any time, t, which is measured in seconds.
If a rock is thrown upward with an initial velocity of 40 feet per second from the top of a 30-foot building, write the height (s) equation using this information, then determine the height after two (2) seconds, how many seconds to reach maximum height and what is maximum height?
:
The three elements of this problem, namely; gravity, initial velocity, and initial height can be represented by a quadratic equation:
:
s = -16t^2 + 40t + 30
:
Substitute 2 sec for t and find s
s = -16(2^2) + 40(2) + 30
s = -64 + 80 + 30
s = 46 ft is the height after 2 sec
:
Find the the time to reach max height by finding the axis of symmetry:
The formula for that is: t = -b/(2a), in this equation b=40, a=-16
t = -40/(2*-16)
t = -40/-32
t = +1.25 sec
:
Find the max height by substituting 1.25 for t in the equation:
s = -16(1.25^2) + 40(1.25) + 30
s = -25 + 50 + 30
s = 55 ft is the max height
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