College Calculus Math

Question

In physics it is shown that the height s, in feet from the ground, of a firework fired upward with an initial velocity of 176 feet per second from a spot on level ground is

s (t) = -16t^2 + 176t,

where t is the elapsed time (in seconds) that the firework is in the air. Find the maximum height of the firework. Prove your answer.

Answer 1

s(t) = -16t^2 +176t

take the derivative and set equal to zero, then solve to time t

v(t)=s'(t) = -32t +176 = 0

t = 176/32 = 5.5 seconds to reach maximum height

plug 5.5 into the original formula for height

max height = s(5.5) = -16(5.5)^2 +176(5.5) = -484 + 968= 484 feet high

that's the differential calculus method

or you could just use algebra and complete the square

-16t^2 +176t

= -16(t^2 - 11t)

= -16(t^2 -11t + 30.25) +16(30.25)

= -16(t- 5.5)^2 + 484, in vertex form of a quadratic parabolic equation with vertex = (5.5, 484) where 5.5 = seconds and 484 = height in feet = maximum height, as the parabola is downward opening