Question

A ball is thrown into the air with an upward velocity of 40 ft/s. Its height h in feet after t seconds is given by the function \large h=-16t^2+40t. What is the ball's maximum height?

Answer 1

`h=25`

Explanation:

From the question we are told that:

Velocity
`v=40fts`

The height h function

`\large h=-16t^2+40t`

Generally time t is mathematically given by

`h'=-32t+40`

`-32t+40=0`

`t=(40)/(32)`

`t=1.25`

Therefore the ball's maximum height is given as

`\large h=-16(1.25)^2+40(1.25)`

`\large h=-16(1.25)^2+40(1.25)`

`h=25`