2. Acceleration Due to Gravity Every object on the earth experiences a common force: the force due to gravity.Every object on the earth experiences a common force: the force due to gravity. This force is always directed toward the center of the earth (downward).This force is always directed toward the center of the earth (downward). The acceleration due to gravity is relatively constant near the Earth’s surface.The acceleration due to gravity is relatively constant near the Earth’s surface. Every object on the earth experiences a common force: the force due to gravity.Every object on the earth experiences a common force: the force due to gravity. This force is always directed toward the center of the earth (downward).This force is always directed toward the center of the earth (downward). The acceleration due to gravity is relatively constant near the Earth’s surface.The acceleration due to gravity is relatively constant near the Earth’s surface. Earth Wg

3. Gravitational Acceleration In a vacuum, all objects fall with same acceleration.In a vacuum, all objects fall with same acceleration. Equations for constant acceleration apply as usual.Equations for constant acceleration apply as usual. Near the Earth’s surface:Near the Earth’s surface: In a vacuum, all objects fall with same acceleration.In a vacuum, all objects fall with same acceleration. Equations for constant acceleration apply as usual.Equations for constant acceleration apply as usual. Near the Earth’s surface:Near the Earth’s surface: a = g = 9.81 m/s 2 Directed downward (usually negative). Directed downward (usually negative).

4. Sign Convention: A Ball Thrown Vertically Upward Velocity is positive (+) or negative (-) based on direction of motion.Velocity is positive (+) or negative (-) based on direction of motion. Displacement is positive (+) or negative (-) based on LOCATION. Displacement is positive (+) or negative (-) based on LOCATION. Release Point UP = + Tippens Acceleration is (+) or (-) based on direction of force (weight).Acceleration is (+) or (-) based on direction of force (weight). y = 0 y = + y = 0 Negative y = - Negative v = + v = 0 v = - Negative v= - Negative a = -

5. Same Problem Solving Strategy Except a = g:  Draw and label sketch of problem.  Indicate + direction and force direction.  List givens and state what is to be found. Given: ____, _____, a = - 9.8 m/s 2 Find: ____, _____   Select equation containing one and not the other of the unknown quantities, and solve for the unknown.

6. Example 7: A ball is thrown vertically upward with an initial velocity of 30 m/s. What are its position and velocity after 2 s, 4 s, and 7 s? Step 1. Draw and label a sketch. a = g + v o = +30 m/s Step 2. Indicate + direction and force direction. Step 3. Given/find info. a = -9.8 ft/s 2 t = 2, 4, 7 s v o = + 30 m/s y = ? v = ?

7. Finding Displacement: a = g + v o = 30 m/s 0 y = (30 m/s)t + ½ (-9.8 m/s 2 )t 2 Substitution of t = 2, 4, and 7 s will give the following values: y = 40.4 m; y = 41.6 m; y = -30.1 m Step 4. Select equation that contains y and not v.

8. Finding Velocity: Step 5. Find v from equation that contains v and not x: Substitute t = 2, 4, and 7 s: v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s a = g + v o = 30 m/s

9. Example 7: (Cont.) Now find the maximum height attained: Displacement is a maximum when the velocity v f is zero. a = g + v o = +96 ft/s To find y max we substitute t = 3.06 s into the general equation for displacement. y = (30 m/s)t + ½ (-9.8 m/s 2 )t 2

10. Example 7: (Cont.) Finding the maximum height: y = (30 m/s)t + ½ (-9.8 m/s 2 )t 2 a = g + v o =+30 m/s t = 3.06 s y = 91.8 m - 45.9 m Omitting units, we obtain: y max = 45.9 m

11. CONCLUSION OF Chapter 6 - Acceleration