x coordinate of positive x intercept

Question

A rocket is launched into the air with an initial velocity of 24 feet per second from a platform 22 feet above the ground. The rocket follows a parabolic pathway. The highest the rocket gets. is 230 feet and that takes 21 seconds. When it hits the ground, 29.9 seconds had passed since the rocket was launched.

If you were to graph the equation of this parabola, what would be the × coordinate of the positive x intercept?

Answer 1

h(t) =-16t^2 +24t + 22

max height is when the derivative h' = 0

= -32t +24

t = 24/32 = 3/4 of a second = 0.75 sec, not 21 seconds

when h(3/4) = 31 feet max

h(3/4) = -16(3/4)^2 +24(3/4) +22 = -9 +18+22 = 31 feet maximum height

yet we're told

max h= 230 feet in 21 seconds

h=0 when t = 29.9 seconds

but h=0 =-16t^2 +24t +22

8t^2 -12t -11 = 0

t =12/16 +/-(1/16)sqr(144+352)

t = 3/4 ++/-sqr496/16

t = about 2.142 and -.642 seconds to hit the ground (ignore the negative square root, as it gives a negative t = -0.64)

NOT 29.9 seconds

there are some mistakes or problems in the problem, possibly miscopied or misprinted

the information is contradictory, to the max

rewrite the quadratic in vertex form

-16t^2 +24t +22

rewrite the quadratic in factor form= about -16(x+.642)(x-2.142)

when x=0 quadratic = 22, not 29.9. 22 is the y intercept, (0, 22), not an x intercept

although -16(x+.65)(x-2.875) with x=0 has 29.9

x coordinate of the x intercept would then be x= 29.9 seconds, a surprising coincidence, although that's a little messy rounding errors

use a graphing calculator, either online or handheld. desmos graphing calculator shows the x zero = 2.142. that's 2.142 seconds to reach ground level, not 29.9 seconds

it's a downward opening parabola with vertex = maximum point = (.75, 31) which means max height of 31 feet reached in 3/4 of a second