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

Answer:

$h=25$

Step-by-step 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

 $h'=-32t+40$

 $-32t+40=0$

 $t=\frac{40}{32}$

 $t=1.25$

Therefore  the ball's maximum height is given as

  $\large h=-16(1.25)^2+40(1.25)$

 $\large h=-16(1.25)^2+40(1.25)$

 $h=25$