Problema Solution
If an object is thrown upward with an initial velocity of 11 ft per second, then its height after t seconds is given by h(t) = -11t^2+44t. What is its maximum height?
Answer provided by our tutors
We need to find the maximum of the quadratic function:
h(t) = -11t^2+44t
Since the quotient in front of t^2 is negative -11 < 0 the function has maximum.
The maximum is calculated by the formula:
h max = c - b^2/(4a), where a = -11, b = 44, c = 0
h max = 0 - 44^2/(4*(-11))
h max = 44 ft
The maximum height is 44 feet.