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A person standing close to the edge on top of a 16​-foot building throws a baseball vertically upward. The quadratic function s (t )equals negative 16 t squared plus 64 t plus 16 models the​ ball's height above the​ ground, s(t), in​ feet, t seconds after it was thrown. After how many seconds does the ball reach its maximum​ height? Round to the nearest tenth of a second if necessary. A. 1.5 seconds

Respuesta :

Answer:

maximum height = 80 feet

It reaches maximum height in 2 seconds

Step-by-step explanation:

A person standing close to the edge on top of a 16​-foot building throws a baseball vertically upward.

[tex]s(t)=-16t^2+64t+16[/tex]

To find maximum height we need to find the vertex

To find x coordinate of vertex use formula

[tex]x=\frac{-b}{2a}[/tex]

a= -16 , b= 64

[tex]x=\frac{-b}{2a}=\frac{-64}{2(-16)} = 2[/tex]

now we plug in 2 for x in s(t) to find maximum height at x=2

[tex]s(t)=-16t^2+64t+16\\s(2)=-16(2)^2+64(2)+16=80[/tex]

so maximum height = 80 feet

It reaches maximum height in 2 seconds

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