ball is projected vertically upwards is at a distance of S ft from the projection point after t seconds , where s=64t-16t^2. will the ball ever be 80 ft above t

no, ball will not go higher than 64ft

Answer:

Ball will never be 80ft above the ground.

Step-by-step explanation:

Consider the provided information.

The provided function is [tex]s=64t-16t^2[/tex]

Here the coefficient of the t² is negative so the maximum will exist at the vertex.

We can find the value of vertex by using the formula: [tex]x =\frac{-b}{2a}[/tex]

Substitute the value of b = 64 and a = -16

[tex]t =\frac{-64}{2(-16)}[/tex]  

[tex]t =2[/tex]

Now substitute the value of t = 2 in the provided function.

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

[tex]s=128-64[/tex]

[tex]s=64[/tex]

Hence, the maximum height can be 64ft.

Ball will never be 80ft above the ground.