A ball is thrown from an initial height of 5 feet with an initial upward velocity of 23/fts. The ball's height h (in feet) after tseconds is given by the following. h=5+23t - 16t2Find all values of t for which the ball's height is 13 feet.

x₂ = 0,59 sec

x₁  = 0,8475 sec

Step-by-step explanation:

h(t) = -16*t² + 23*t + 5

h(t) is the trajectory of the ball, the curve is a parable opens downwards

if we force h(t) = 13 feet, we get;

h(t) =  13

13 =  -16*t² + 23*t + 5   ⇒  -16*t² + 23*t - 8 = 0

or    16*t² - 23*t + 8 = 0

The above expression is a second degree equation, we proceed to solve it for t

x =  [ 23  ±  √529 - 512 ] /32

x = [ 23 ± √17 ] /32

x₁ =  [ 23 + 4,12 ]/32    ⇒  x₁  = 27,12/32    ⇒   x₁  = 0,8475 sec

x₂ =   [ 23 - 4,12 ]/32   ⇒  x₂ = 18,88 /32   ⇒   x₂ = 0,59 sec

the final function for the graph becomes x-2=-1. before graphing you solve the equation, where x=1. then graph accordingly.   ( on the x-axis, 1 on the line will possess a vertical line.)

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer