### Home > APCALC > Chapter 6 > Lesson 6.1.3 > Problem6-39

6-39.

VERTICAL MOTION

The acceleration due to gravity is a constant near the surface of the Earth. Its accepted values are $-32 \text{ ft}/\text{s}^2$ or $–9.8 \text{ m}/\text{s}^2$. These values are all the information needed to derive the equations of $v(t)$ and $a(t)$ for vertical motion for any dropped or thrown object. Homework Help ✎

1. Describe how can you find a velocity function when given the acceleration function.

Acceleration is the rate of change of velocity.

2. Write an equation for velocity, $v(t)$, if $a(t) =-32 \text{ ft}/\text{s}^2$ and if the initial velocity of an object, $v_0$, is $120$ feet per second.

There is a relationship between the +C and the initial velocity.

3. Write an equation for height, $s(t)$, if the starting position of the object, $x_0$, is $100$ feet above ground.

Velocity is the rate of change of position.

s(t) = −16t2 + 120t + 100

4. What is the velocity of the object when it hits the ground

First find the TIME when the object hits the ground: s(t) = 0.Then evaluate v(t) at that time.

5. What is the maximum height attained by the object?

Find the vertex of the position function. You could use Algebra or Calculus.

6. At what time is the object’s speed greatest?

Where is the slope of s(t) the steepest? Remember that speed is the absolute value of velocity.