Question

If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a function of time.

Given by s= -8t^2 + 32t. what is the maximum height reached by the ball?

Answer 1

Maximum height reached by the ball is 32 meters.

Step-by-step explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

`s=-8t^2+32t`...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

`(ds)/(dt)=0`

`(d(-8t^2+32t))/(dt)=0`

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

`s=-8(2)^2+32(2)`

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.