this is a very hard question

Question

A ball is thrown from an initial height of 2 feet with an initial upward velocity of 23ft/s.The ball height h(in feet) after t seconds is given by the following h=2+23+-16t^2 find all values of t for which the ball height is 10 feet

Answer 1

h=10 when t= .59 seconds and .85 seconds, rounded off to 2 decimal places

h(t) = -16t^2 + 23t +2 where h(t) equals the height, -16t^2 is the deceleration due to gravity. 23t is the velocity and 2 is the initial height.

set h(t) = 10 and solve for t

-16t^2 +23t +2 = 10

-16t^2 +23t -8 = 0

use the quadratic formula to solve for t

t = (-23+ or - sqr(23^2 -4(-16)(-8))/-32 = 23/32 + or - (sqr17)/-32 = about .72 + or - .13 = .85, .59 seconds

at about .72 seconds the ball reaches a maximum height.

at plus & minus .13 seconds, from .72 seconds it reaches 10 feet.

maximum height is -16(.72)^2+23(.72)+2 = about 10.3 feet. About 1/10 of a second before and after the time for maximum height, the ball reaches 10 feet in height.