an object is projected vertically upward from the top of a building with an initial velocity of 128 ft/sec…

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.

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.

Answer

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