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Free Fall and Acceleration Due to Gravity: Understanding Falling Objects on Earth

Explore the concepts of free fall and the acceleration due to gravity on Earth. Learn about position-time and velocity-time graphs, the constant acceleration of free fall, and how to calculate velocities and maximum heights of falling objects.

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Free Fall and Acceleration Due to Gravity: Understanding Falling Objects on Earth

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  1. Chapter 2b Falling Objects

  2. g W Earth 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.

  3. Free-falling objects do not encounter air resistance. All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s2). Free Fall and the Acceleration of Gravity

  4. Position-Time Graph Velocity-Time Graph Graphs of Free Fall

  5. 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 = -9.80 m/s2 Directed downward (usually negative).

  6. + a = g Example 7:A ball is thrown vertically upward with an initial velocity of 30 m/s. a.) What is its velocity after 2 s, 4 s, and 7 s?b.) What is it’s maximum height? Step 1. Draw and label a sketch. Step 2. Indicate + direction and force direction. Step 3. Given/find info. a = -9.8 m/s2 t = 2, 4, 7 s vo = + 30m/sy = ?v = ? vo= +30 m/s

  7. + a = g vo= 30 m/s Finding Velocity: Step 5. Find vffrom equation that contains vand not x: Substitute t = 2, 4, and 7 s: v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s NOTE the direction of the velocity!

  8. + a = g vo = +96 ft/s Example 7: (Cont.) Now find the maximum height attained: Displacement is a maximum when the velocity vf is zero. To find ymaxwe substitute t = 3.06 s into the general equation for displacement. y = (30 m/s)t + ½(-9.8 m/s2)t2

  9. + a = g vo =+30 m/s 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: y = (30m/s)(3.0s) +1/2(-9.8m/s2)(3.06s)2 y = 91.8 m - 45.9 m ymax = 45.9 m

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