If a ball is thrown vertically upward with an initial velocity of $128$ ft$/$s$,$ then its height after t seconds is

s $=128$t $16$t$2.$

$($a$)$

What is the maximum height $($in ft$)$ reached by the ball?

$($b$)$

What is the velocity $($in ft$/$s$)$ of the ball when it is $240$ ft above the ground on its way up$?$ On its way down?

Step $1$

$($a$)$What is the maximum height $($in ft$)$ reached by the ball?

In the described situation, we note that after the ball is thrown upwards, the velocity will eventually slow to $0$ ft$/$s as it reaches its maximum height just before it starts to fall. Therefore, to find the maximum height reached by the ball, we need to first determine at what time t the velocity of the ball is equal to $0$ ft$/$s$.$

Recall that if the position of an object after t seconds is given by an equation

s$($t$)=128$t $16$t$2,$

then the instantaneous velocity

v$($t$)$

is determined by the following relationship.

v$($t$)=$ s$($t$)$

Therefore, we must first find

s$($t$).$

Doing so gives the following result.

s$($t$)=128$t $16$t$2$s$($t$)=$