Acceleration Due to Gravity EQ: How is the motion of an object moving vertically different from one moving horizontally? EQ: How is the motion of an object moving vertically different from one moving horizontally?

Acceleration Due to Gravity In physics, a uniformly accelerated motion in a vertical line is called free fall. 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). Free fall can be analyzed using the same kinematic equations we used to analyze horizontal motion. But there are a few factors we must take into consideration. In physics, a uniformly accelerated motion in a vertical line is called free fall. 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). Free fall can be analyzed using the same kinematic equations we used to analyze horizontal motion. But there are a few factors we must take into consideration.

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

The Acceleration of Gravity (g) If two rocks with different mass are dropped (v i = 0) from the same height at the same time, would one land before the other? Or would they land at the same time? Galileo demonstrated that g is the same for all objects, regardless of their mass! This was confirmed by the Apollo astronauts on the Moon, where there is no air resistance 35ZD1Ig 35ZD1Ig

Falling objects accelerate at a constant rate (Galileo): All objects (regardless of their mass) experience the same acceleration when free falling. No air resistance present. When the only force is gravity, the acceleration is the same for all objects. On Earth, this acceleration (g) is 9.8 m/s 2. When an object is dropped or simple falls, it will have an v i = 0 m/s since no initial force was applied to make it move.

Gravitational Acceleration When an object is thrown straight up, it will have a velocity of 0 m/s 2 at its peak height just before it begins its free fall. – This means that for a split second, an object will stop in mid air while it transitions from upward motion to downward motion. Because gravity is constant, the rate at which an object decelerates as it moves upward, is the same rate it accelerates at as it falls downward. – In other words, an object falling will have the same speed at any point that it did on the way up. 6

Various Types of Free Fall 7 0 m/s 9.8 m/s 19.6 m/s 29.4 m/s 39.2 m/s 49.0 m/s

Graphing Free Fall Motion

Talk out your misconceptions about free falling objects with a shoulder partner then at the bottom of your notes explain: - Why any 2 objects will hit the ground first regardless of their mass (in the absence of air resistance) -Why the initial velocity of free falling objects is zero -What direction does gravity act on (side to side or up and down) -What does it mean to say that the acceleration of gravity is 9.80 m/s 2 (for ex: what is happening to the object every second as it falls)

Sign Convention: A Ball Thrown Vertically Upward Velocity is positive (+) or negative (-) based on direction of motion. Displacement is positive (+) or negative (-) based on LOCATION. Release Point UP = + Acceleration is gravity which is always downward (negative). x= 0 v= - x= + v= + x= + v= 0 x= + v= - x= - v= -

Kinematic Equations Horizontal Motion v f = v i + g(Δt) Δy = ½ (v i + v f )Δt Δy = v i (Δt) + ½ g(Δt) 2 v f 2 = v i 2 + 2gΔy Drop/fall  V i =0 m/s Height, tall  Δy g= acceleration due to gravity = m/s 2 Vertical Motion v f = v i + a(Δt) Δx = ½ (v i + v f )Δt Δx = v i (Δt) + ½ a(Δt) 2 v f 2 = v i 2 + 2aΔx

Same Problem Solving Strategy Except a = g:  Draw and label sketch of problem.  Indicate + direction.  List givens and state what is to be found.  Select equation containing one and not the other of the unknown quantities, and solve for the unknown.  Don’t Forget Units!!!!!

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 i = +30 m/s Step 2. Indicate + direction. Step 3. Identify given variables. a = -9.8 ft/s 2 t = 2, 4, 7 s v i = + 30 m/s y= ? v = ?

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

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

Example 7: (Cont.) Now find the maximum height attained: Now find an equation that uses y but not t. Keep in mind the velocity at its highest point is zero a = g + v o = +30 m/s

The tallest Sequoia sempervirens tree in California’s Redwood National Park is 111 m tall. Suppose an object is thrown downward from the top of that tree with a certain initial velocity. If the object reaches the ground in 3.80 s, what is the object’s initial velocity?

Kinematic Equations Horizontal Motion v f = v i + g(Δt) Δy = ½ (v i + v f )Δt Δy = v i (Δt) + ½ g(Δt) 2 v f 2 = v i 2 + 2gΔy Drop/fall  V i =0 m/s Height, tall  Δy g= acceleration due to gravity = m/s 2 Vertical Motion v f = v i + a(Δt) Δx = ½ (v i + v f )Δt Δx = v i (Δt) + ½ a(Δt) 2 v f 2 = v i 2 + 2aΔx

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