 # 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

Math

## Question

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 the ground

• ### 1. User Answers Anonym

no, ball will not go higher than 64ft

• ### 2. User Answers FelisFelis

Ball will never be 80ft above the ground.

Step-by-step explanation:

Consider the provided information.

The provided function is $$s=64t-16t^2$$

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: $$x =\frac{-b}{2a}$$

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

$$t =\frac{-64}{2(-16)}$$

$$t =2$$

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

$$s=64(2)-16(2)^2$$

$$s=128-64$$

$$s=64$$

Hence, the maximum height can be 64ft.

Ball will never be 80ft above the ground.