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

Answer 1

no, ball will not go higher than 64ft

Answer 2

Answer:

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.