Question 135345
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