#### To determine

**To find:** The maximum height of the ball if the ball is thrown with the velocity 80 ft/s and with the height after t seconds is s=80t−16t2.

#### Answer

The ball reaches the maximum height of 100 ft.

#### Explanation

The height of the ball is s=80t−16t2.

Find the derivative of s(t) and obtain the velocity.

v(t)=s'(t)=80−32t

Thus, the velocity is v(t)=80−32t.

Recall the fact that the velocity of the ball is 0 ft/s at the maximum height.

Set v(t)=0 and obtain the value of *t*.

80−32t=032t=80t=52t=2.5

Thus, the ball reaches the maximum height at t = 2.5 ft/s.

Substitute *t* = 2.5 in s(t) and obtain the maximum height.

s(2.5)=80(2.5)−16(2.5)2=200−100=100

Thus, the ball reaches it maximum height of 100 ft/s.

#### To determine

**To find:** The velocities of the ball when the ball is 96 ft above the ground on its way up and on its way down.

#### Answer

The velocity of the ball when is 96 ft above the ground when its way up is 16 ft/s.

The velocity of the ball when is 96 ft above the ground when its way down is −16 ft/s.

#### Explanation

**Given:**

The ball is thrown with the velocity of 80 ft/s and with the height after t seconds is s=80t−16t2.

**Calculation:**

Set s(t)=96 and obtain the value of *t*.

80t−16t2=9616t2−80t+96=016(t2−5t+6)=016(t−3)(t−2)=0

Thus, the ball reaches 96 ft on its way up at *t* = 2 s and on its way down at *t* = 3 s.

From part (1), v(t)=80−32t.

Substitute *t* = 2 for v(t) and obtain the velocity on its way up.

v(2)=80−32(2)=80−64=16

Thus, the velocity of the ball when it reaches 96 ft above the ground level on its way up is 16 ft/s.

Substitute *t* = 3 for v(t) and obtain the velocity on its way up.

v(3)=80−32(3)=80−96=−16

Thus, the velocity of the ball when it reaches 96 ft above the ground level on its way down is −16 ft/s.