Quadratic Equations

Question

April shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high. The pathway of the arrow can be represented by the equation h = -16t ² + 80t + 25

Answer 1

take the derivative and set = 0, solve for t to find time at maximum height

-32t + 80 = 0

t = 80/32 = 5/2 = 2.5 seconds to reach max height

plug 2.5 into the original h(t) equation to solve for that maximum height

h(2.5) = -16(2.5)^2 + 80(2.5) + 25 = (-16(2.5)+80)(2.5 + 25 = 40(2.5)+25 =115 ft.

h(2.5) = 115 feet, max height.

if you want to know when it hits the ground set h(t) = 0 and solve for t

-16t^2 + 80t + 25 =0

use the quadratic equation if it doesn't factor, or complete the square

ignore the negative square root

t = 80/32 or 2.5 + (1/32)sqr(80^2-4(16)(25) = about 2.5+2.8 = 5.3 seconds to reach the ground

that's the usual questions on parabolic equations with height, gravity, initial velocity and height