Acceleration Due to Acceleration Due to GravityGravity

• Every object on the earth Every object on the earth experiences a common force: experiences a common force: the force due to gravity.the force due to gravity.

• This force is always directed This force is always directed toward the center of the toward the center of the earth (downward).earth (downward).

• The acceleration due to The acceleration due to gravity is relatively constant gravity is relatively constant near the Earth’s surface.near the Earth’s surface.

• Every object on the earth Every object on the earth experiences a common force: experiences a common force: the force due to gravity.the force due to gravity.

• This force is always directed This force is always directed toward the center of the toward the center of the earth (downward).earth (downward).

• The acceleration due to The acceleration due to gravity is relatively constant gravity is relatively constant near the Earth’s surface.near the Earth’s surface.

Earth

Wg

Gravitational AccelerationGravitational Acceleration

• In a vacuum, all objects fall In a vacuum, all objects fall with same acceleration.with same acceleration.

• Equations for constant Equations for constant acceleration apply as acceleration apply as usual.usual.

• Near the Earth’s surface:Near the Earth’s surface:

• In a vacuum, all objects fall In a vacuum, all objects fall with same acceleration.with same acceleration.

• Equations for constant Equations for constant acceleration apply as acceleration apply as usual.usual.

• Near the Earth’s surface:Near the Earth’s surface:

aa = g = = g = 9.81 m/s9.81 m/s22

Directed downward (usually Directed downward (usually negative).negative).

Sign Convention:Sign Convention: A Ball Thrown A Ball Thrown

Vertically Vertically UpwardUpward

• Velocity is positive (+) Velocity is positive (+) or negative (-) based or negative (-) based on on direction of motiondirection of motion..

• Velocity is positive (+) Velocity is positive (+) or negative (-) based or negative (-) based on on direction of motiondirection of motion..

• Displacement is positive Displacement is positive (+) or negative (-) (+) or negative (-) based on based on LOCATIONLOCATION. .

• Displacement is positive Displacement is positive (+) or negative (-) (+) or negative (-) based on based on LOCATIONLOCATION. .

Release Point

UP = +

TippensTippens

• Acceleration is (+) or (-) Acceleration is (+) or (-) based on direction of based on direction of forceforce (weight). (weight).

y = 0

y = +

y = +

y = +

y = 0

y = -NegativeNegative

v = +

v = 0

v = -

v = -

v= -NegativeNegative

a = -

a = -

a = -

a = -

a = -a = -

Same Problem Solving Same Problem Solving Strategy Except Strategy Except aa = g = g::

Draw and label sketch of problem.Draw and label sketch of problem.

Indicate Indicate ++ direction and direction and forceforce direction. direction.

List givens and state what is to be found.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:Example 7: A ball is thrown vertically A ball is thrown vertically upward with an initial velocity of upward with an initial velocity of 30 m/s30 m/s. . What are its position and velocity after What are its position and velocity after 2 s2 s, , 4 s4 s, and , and 7 s7 s??

Step 1. Draw and label a sketch.

a = g

+

vo = +30 m/s

Step 2. Indicate + direction and force direction.Step 3. Given/find info.

a = -9.8 ft/s2 t = 2, 4, 7 s

vo = + 30 m/s y = ? v = ?

Finding Finding Displacement:Displacement:

a = g

+

vo = 30 m/s

0

y = y = (30 m/s)(30 m/s)tt + + ½½(-9.8 (-9.8 m/sm/s22))tt22

Substitution of t = 2, 4, and Substitution of t = 2, 4, and 7 s will give the following 7 s will give the following values: values:

y = 40.4 m; y = 41.6 m; y = -30.1 my = 40.4 m; y = 41.6 m; y = -30.1 m

210 0 2y y v t at

Step 4. Select equation that contains y and not v.

Finding Velocity:Finding Velocity:Step 5. Find v from equation that contains v and not x:

Step 5. Find v from equation that contains v and not x:

Substitute t = 2, 4, and 7 Substitute t = 2, 4, and 7 s:s:

v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/sv = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s

a = g

+

vo = 30 m/s

0fv v at 0fv v at

230 m/s ( 9.8 m/s )fv t

Example 7: (Cont.)Example 7: (Cont.) Now Now find the find the maximum heightmaximum height attained:attained:

Displacement is a Displacement is a maximum when the maximum when the velocity velocity vvff is zero. is zero. a =

g

+

vo = +96 ft/s

230 m/s ( 9.8 m/s ) 0fv t

2

30 m/s; 3.06 s

9.8 m/st t

To find To find yymaxmax we we substitute substitute tt = 3.06 s = 3.06 s into the general into the general equation for equation for displacement.displacement.y = y = (30 m/s)(30 m/s)tt + + ½½(-9.8 (-9.8

m/sm/s22))tt22

Example 7: (Cont.)Example 7: (Cont.) Finding the Finding the maximum maximum height:height:

y = y = (30 m/s)(30 m/s)tt + + ½½(-9.8 (-9.8 m/sm/s22))tt22

a = g+

vo =+30 m/s

tt = 3.06 = 3.06 ss

212(30)(3.06) ( 9.8)(3.06)y

yy = 91.8 m - 45.9 = 91.8 m - 45.9 mm

Omitting units, we obtain:Omitting units, we obtain:

ymax = 45.9 m

CONCLUSION OF CONCLUSION OF Chapter 6 - AccelerationChapter 6 - Acceleration