Quadratic equations and functions

Question

A ball is thrown from an initial of 1 meter with an initial upward velocity of 30 m/s. The balls height h (in meters) after t seconds is given by the following

Answer 1

h(t) = -4.9t^2 +30t +1

maximum height is reached when

h'= -9.8t +30 = 0

t = 30/9.8 = 150/49 = 3 3/49 seconds

= about 3 seconds

max height = h(150/49) = -4.9(150/49)^2 + 30(150/49) +1

= -4.9(9.1837)+ 4500/49 +1

= -45+ 91 41/49+1

= 47 41/49 meters high

it hits the ground when

h(t) = 0 = -4.9t^2 +30t + 1 = 0

4.9t^2 -30t -1 = 0

use the quadratic formula to solve for t

t =30/9.8 +(1/9.8)sqr(30^2 +4(4.9))

t = (30 +sqr919.3)/9.8

= 60.32/9.8= 1508/245=6 38/235 seconds

roughly twice the time to reach max height

= about 2(3) =

about 6 seconds

standard height formula

is h(t) = at^2/2 + vot + ho

where ho = initial height = 1 meter

vo = initial height = 30 m/s

a = -9.8 m/s^2 = acceleration due to gravity, at sea level

h= -4.9t^2 +30t +1 is a downward opening parabola with vertex = max height

= (h,k) where h = time to reach max height and k = max height, in meters

= -4.9(t^2 -30t/4.9) + 1

=-4.9(t^2 -30t/4.9+ (30/9.8)^2) + 1 +4.9(30/9.8)^2

= -4.9(t-30/9.8)^2 + (19.6+ 30^2)/19.6

=-4.9(t-30/9.8)^2 + 919.6/19.6 is the quadratic in vertex form

vertex = (h,k) = (30/9.8, 919.6/19.6)

= about (3,47) where h= 3 seconds and k = 47meters high at the highest point