Question

A football player throws a ball down the side line. The ball's height above the

ground is modeled by the equation h=-8t^2+24t+4, where h is the height
in feet and t is the time in seconds.
At what two times will the ball be at a height of 16 feet?

Answer 1

Answer: t₁=1.5-0.5√3 c ≈0.634 c t₂=1.5+0.5√3 c ≈2.366 c

Explanation:

h=-8t²+24t+4 h=16 feet t₂=?

`16=-8t^2+24t+4\\\\16-16=-8t^2+24t+4-16\\\\0=-8t^2+24t-12\\`

Divide both parts of the equation by -4:

`0=2t^2-6t+3`

Thus,

`2t^2-6t+3=0`

a=2 b=-6 c=3

D=b²-4ac

D=(-6)²-4(2)(3)

D=36-24

D=12

√D=√12

√D=√(4*3)

√D=2√3

`t=(-bб√(D) )/(2a) \\\\t=(-(-6)б2√(3) )/(2(2)) \\\\t=(6б2√(3) )/(4) \\\\t_1=1.5-0.5√(3)\ c \\\\t_2=1.5+0.5√(3)\ c`