Chapter 2 Motion in One Dimension Ms. Hanan Anabusi

2-3 Falling Objects Objectives: Relate the motion of a freely falling body to motion with constant acceleration. Calculate displacement, velocity, and time at various points in the motion of a freely falling object. Compare the motion of different objects in free fall.

2-3 Falling Objects Vocabulary: Free fall Gravity

Falling Objects Free fall is the motion of an object falling with a constant acceleration (in the absence of air resistance) [see Fig 2-14]. Free fall acceleration is denoted by the symbol g. At the surface of Earth the magnitude of g is 9.81 m/s2. (This is an average, used in calculations.) This acceleration is downward, toward the center of the earth, therefore the acceleration of objects in free fall (on Earth) is: a = -g = -9.81m/s2

Falling Objects An object thrown in the air will continue to move upward for some time, stop momentarily at the peak, change direction, and begin to fall. Acceleration due to gravity (g) is always acting on an object moving through the air and the value is always -9.81m/s2. As the object moves up it has a positive velocity. At the peak the object has velocity = 0 m/s. As the object moves down it has a negative velocity.

Sample problem F page 63 vi=6.0m/s a=-9.81m/s2 Dy=-2.0m Jason hits a volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. If the volleyball starts from 2.0m above the floor, how long will it be in the air before it strikes the floor? Assume that Jason is the last player to touch the ball before it hits the floor. vi=6.0m/s a=-9.81m/s2 Dy=-2.0m vf2 = vi2 + 2aDy vf = vi + aDt vf2= (6.0m/s)2 + 2(-9.81m/s2)(-2.0m) = 75m2/s2 vf = +/-8.7m/s = -8.7m/s (because the ball is moving down) Solve for Dt: Dt=(vf - vi)/a = (-8.7m/s - 6.0m/s)/-9.81m/s2 = -14.7 m/s / -9.81 m/s2 = 1.50s Solve Practice E #1-6 page 58

Velocity and acceleration Motion + Speeding up - Slowing down + or - Constant velocity Speeding up from rest Remaining at rest

Sample problem F page 63 vi=6.0m/s a=-9.81m/s2 Dy=-2.0m Dt= ? vf= ? Jason hits a volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. If the volleyball starts from 2.0m above the floor, how long will it be in the air before it strikes the floor? Assume that Jason is the last player to touch the ball before it hits the floor. vi=6.0m/s a=-9.81m/s2 Dy=-2.0m Dt= ? vf= ? vf2 = vi2 + 2aDy vf = vi + aDt vf2= (6.0m/s)2 + 2(-9.81m/s2)(-2.0m) = 75m2/s2 vf = +/-8.7m/s = -8.7m/s (because the ball is moving down) Solve for Dt: Dt=(vf - vi)/a = (-8.7m/s - 6.0m/s)/-9.81m/s2 = -14.7 m/s / -9.81 m/s2 = 1.50s

Class work: Homework: Solve Practice F #2, and 4 on page 64 Chapter review, page 71-72, questions 41 and 43

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