A baseball player hits a ball and it goes straight up into t

A baseball player hits a ball and it goes straight up into the air (fly ball). The height of the ball is described as:

h(t)=-16t^2+42t+10

When is the ball at the maximum height? (Hint: the maximum height occurs at the vertex)

Solution

given h(t)=-16t²+42t+10

h(t)=(-16*(t²-(42_/16)t))+10

h(t)=(-16*(t²-(21_/8)t))+10

h(t)=(-16*(t²-(21_/8)t+(21_/16)²-(21_/16)²))+10

h(t)=(-16*(t²-(21_/8)t+(21_/16)²)) +(16*(21_/16)²)+10

h(t)=(-16*(t²-(21_/16)t)²) +(441_/16)+10

h(t)=(-16*(t²-(21_/16)t)²) +(601_/16)

vertex is (21_/16 ,601_/16)

maximum height =601_/16=37.5625

ball is at maximum height when t =21_/16=1.3125