Question

an object is projected vertically upward from the top of a building with an initial velocity of 128 ft/sec. its distance s(t) in feet above the ground after t seconds is given by the equation s(t)= - 16t^2 + 128t + 90. (a) find its maximum distance above the ground. (b) find the height of the building.

Explanation:

Step1: Identify the coefficients of the quadratic function

The height - function is \(s(t)=-16t^{2}+128t + 90\), where \(a=-16\), \(b = 128\), and \(c = 90\). The vertex of a quadratic function \(y = ax^{2}+bx + c\) has its \(t\) - coordinate given by \(t=-\frac{b}{2a}\).

Step2: Calculate the time \(t\) at which the maximum height occurs

Substitute \(a=-16\) and \(b = 128\) into the formula \(t=-\frac{b}{2a}\).
\[t=-\frac{128}{2\times(-16)}=\frac{-128}{-32}=4\]

Step3: Find the maximum height

Substitute \(t = 4\) into the function \(s(t)=-16t^{2}+128t + 90\).
\[s(4)=-16\times4^{2}+128\times4 + 90=-16\times16+512 + 90=-256+512 + 90=346\]

Step4: Find the height of the building

The height of the building is the height at \(t = 0\). Substitute \(t = 0\) into \(s(t)\).
\[s(0)=-16\times0^{2}+128\times0 + 90=90\]

Answer:

(a) 346 ft
(b) 90 ft