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

6-39.
1. VERTICAL MOTION

2. The acceleration due to gravity is a constant near the surface of the Earth. Its accepted values are −32.2 ft/sec2 or −9.8 m/sec2 . These values are all the information needed to derive the equations 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.

2. Find v(t) if a(t) = −32 ft/sec2 and if the initial velocity of an object, v0 , is 120 feet per second.

3. Find its height s(t) if the starting position of the object, x0 , is 100 feet above ground.

4. Find the velocity of the object when it hits the ground.

5. Find the maximum height attained by the object.

6. At what time is the object's speed greatest?  Acceleration is the rate of change of velocity... There is a relationship between the +C and the initial velocity. Velocity is the rate of change of position...

s(t) = −16t2 + 120t + 100 First find the TIME when the object hits the ground: s(t) = 0.Then evaluate v(t) at that time. Find the vertex of the position function. You could use Algebra or Calculus. Where is the slope of s(t) the steepest? Remember that speed is the absolute value of velocity.