Question 1183267


A golf ball is hit from the top of a tee. The quadratic equation 
{{{y=-5x^2 + 20x + 0.05 }}}

describes its height, {{{y}}}, in metres as time,{{{x}}}, in seconds passes. 

Determine how long the ball is in the air. 

First, you have to determine how long it will take for the golf ball to reach its maximum height.

so, rewrite equation in vertex form

{{{y=a(x-h)^2+k}}} where {{{h}}} and {{{k}}} are coordinates of vertex

{{{y=-5x^2 + 20x + 0.05 }}}....group
{{{y=(-5x^2 + 20x) + 0.05 }}}....factor out {{{-5}}}
{{{y=-5(x^2 -4x) + 0.05 }}}
{{{y=-5(x^2 -4x+b^2) -(-5b^2)+ 0.05 }}}....{{{b=4/2=2}}}
{{{y=-5(x^2 -4x+2^2) +5*2^2+ 0.05 }}}
{{{y=-5(x -2)^2 +20+ 0.05 }}}
{{{y=-5(x -2)^2 +20.05 }}}

=> {{{h=2}}} and {{{k=20.05}}}
to reach its maximum height of {{{20.05}}}, it will take {{{2}}} seconds

The time in the air is twice the one I found solving for t where the height is maximum.

the ball is {{{4}}} seconds in the air