#### To determine

**To find:** The velocity of the rock after 2 s if a rock is thrown with the velocity of 15 m/s and the height of the rock after *t* seconds is h(t)=15t−1.86t2.

#### Answer

The velocity after 2 s is 7.56 m/s.

#### Explanation

The height of the rock is h(t)=15t−1.86t2.

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

v(t)=h'(t)=15−3.72t

Thus, the velocity is v(t)=15−3.72t (1)

Substitute *t* = 2 in equation (1) and obtain the velocity after 2 s.

v(2)=15−3.72(2)=15−7.44=7.56

Thus, the velocity after 2 s is 7.56 m/s.

#### To determine

**To find:** The velocities of the rock when it reaches 25 m on its way up and on its way down.

#### Answer

The velocity of the rock when its height is 25 m on its way up is 6.24 m/s.

The velocity of the rock when its height is 25 m on its way down is −6.24 m/s.

#### Explanation

**Given:**

The rock is thrown with the velocity of 15 m/s and the height of the rock after *t* seconds is h(t)=15t−1.86t2.

**Calculation:**

Set h(t)=25 and obtain the value of *t*.

15t−1.86t2=251.86t2−15t+25=0

Use quadratic formula and simplify further.

t=15±152−4(1.86)(25)2(1.86)=15±225−1863.72=15±493.72=15±73.72

Therefore, the solutions are t≈2.35 and t≈5.71.

Thus, the ball reaches 25 ft on its way up at *t* = 2.35 s and on its way down at *t* = 5.71 s.

From part (1), v(t)=15−3.72t.

Substitute *t* = 2.35 for v(t) and obtain the velocity on its way up.

v(2.35)=15−3.72(2.35)=15−8.742≈6.24

Thus, the velocity of the rock when it reaches 25 m on its way up is 6.24 m/s.

Substitute *t* = 5.71 for v(t) and obtain the velocity on its way up.

v(5.71)=15−3.72(5.71)=15−21.2412≈−6.24

Thus, the velocity of the rock when it reaches 25 m on its way down is −6.24 m/s.