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Acceleration 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). • The acceleration due to gravity is relatively constant near the Earth’s surface. g W Earth Gravitational Acceleration • In a vacuum, all objects fall with same acceleration. • Equations for constant acceleration apply as usual. • Near the Earth’s surface: a = g = 9.81 m/s2 Directed downward (usually negative). Sign Convention: A Ball Thrown Vertically Upward avy== =-0 + avy= ==-++ ya =+-v == • UP = + Release Point Displacement is positive (+) or negative (-) based on LOCATION. vya== =-0- • yv= =a=Negative Negative Tippens Velocity is positive (+) or negative (-) based on direction of motion. • Acceleration is (+) or (-) based on direction of force (weight). 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/s2 Find: ____, _____ Select equation containing one and not the other of the unknown quantities, and solve for the unknown. 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. Step 2. Indicate + direction and force direction. + a=g Step 3. Given/find info. a = -9.8 ft/s2 t = 2, 4, 7 s vo = + 30 m/s y = ? v = ? vo = +30 m/s Finding Displacement: Step 4. Select equation that contains y and not v. 0 y y0 v0t at 1 2 + a=g 2 y = (30 m/s)t + ½(-9.8 m/s2)t2 Substitution of t = 2, 4, and 7 s will give the following values: vo = 30 m/s y = 40.4 m; y = 41.6 m; y = -30.1 m Finding Velocity: Step 5. Find v from equation that contains v and not x: v f v0 at v f 30 m/s (9.8 m/s )t + a=g 2 Substitute t = 2, 4, and 7 s: vo = 30 m/s v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s Example 7: (Cont.) Now find the maximum height attained: Displacement is a maximum when the velocity vf is zero. v f 30 m/s (9.8 m/s )t 0 2 30 m/s t ; t 3.06 s 2 9.8 m/s To find ymax we substitute t = 3.06 s into the general equation for displacement. + a=g vo = +96 ft/s y = (30 m/s)t + ½(-9.8 m/s2)t2 Example 7: (Cont.) Finding the maximum height: y = (30 m/s)t + ½(-9.8 m/s2)t2 t = 3.06 s Omitting units, we obtain: + a=g y (30)(3.06) (9.8)(3.06) 1 2 y = 91.8 m - 45.9 m vo =+30 m/s ymax = 45.9 m 2 CONCLUSION OF Chapter 6 - Acceleration