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## Freely Falling Objects

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**Freely Falling Objects**• Important & common special case of uniformly accelerated motion: “FREE FALL” Objects falling in Earth’s gravity. Neglect air resistance. Use one dimensional uniform acceleration equations (with some changes in notation, as we will see)**Near the surface of the Earth, all objects experience**approximately the same acceleration due to gravity. This is one of the most common examples of motion with constant acceleration. In the absence of air resistance, all objects fall with the same acceleration, although this may be tricky to tell by testing in an environment where there is air resistance.**Falling Objects**• Experiment: • Ball & light piece of paper dropped at the same time. Repeated with wadded up paper.**The acceleration due to gravity at the Earth’s surface is**Approximately 9.80 m/s2. At a given location on the Earth & in the absence of air resistance, all objects fall with the same constant acceleration. • Experiment: • Rock & feather dropped at the same time in air. Repeated in vacuum.**Experiment finds that the acceleration of falling objects**(neglecting air resistance) is always (approximately) the same, no matter how light or heavy the object. • The magnitude of the acceleration due to gravity, a g g = 9.8 m/s2(approximately!)**The acceleration of falling objects is always the same, no**matter how light or heavy. • Acceleration due to gravity, g = 9.8 m/s2 • First proven by Galileo Galilei A Legend: He dropped objects off of the leaning tower of Pisa.**The magnitude of the acceleration due to gravity: g = 9.8**m/s2(approximately!) • It has a slight dependence on the location on Earth, on the latitude & the altitude:**The Algebraic Sign (+ or - ?) in the**Kinematics Equations of the One-Dimensional Vector Gravitational Acceleration • Note: My treatment is slightly different than the book’s, but it is equivalent! • To treat motion of falling objects, we use the same equations we already have, but change notation slightly: Replace a by g = 9.8 m BUT in the equations, it could have a + or a - sign in front of it! We discuss this next! • Usually, we consider vertical motion to be in the y direction & so replace x by y and x0 by y0 (often y0 = 0)**NOTE!!!**Whenever I (or the author!) write the symbol g, it ALWAYS means the POSITIVE numerical value 9.8 m/s2!! It is NEVERnegative!!! The sign (+ or -) of the one-dimensional gravitational acceleration VECTOR is taken into account in the Equationswe now discuss!**The Sign of g in the 1d Equations**• The magnitude (size) of g = 9.8 m/s2(POSITIVE!) • But, acceleration is a (1 dimensional) VECTOR with 2 possible directions. • Call these + and -. • However, which way is + and which way is – is ARBITRARY& UP TO US! • It may seem “natural” for “up” to be + y & “down” to be - y, but we could also choose (we sometimes will!) “down” to be + y and “up” to be - y So, in the equations g could have a + or a - sign in front of it, depending on our choice!**Directions of Velocity & Acceleration**• Objects in free fall ALWAYShave DOWNWARD acceleration. • We still use the same equations for objects thrownupwardwith some initial velocity v0 • An object goes up until it stops at some point & then it falls back down. The acceleration vector isalwaysgin thedownwarddirection. For the first half of flight, the velocity isUPWARD. For the first part of the flight, velocity & acceleration are inopposite directions!**VELOCITY & ACCELERATION**ARE NOT NECESSARILYIN THE SAME DIRECTION!**Equations for Objects in Free Fall**• Written taking “up” as + y v = v0 - gt (1) y = y0 + v0t - (½)gt2 (2) (v)2 = (v0)2 - 2g(y - y0) (3) vavg = (½)(v + v0) (4) g = 9.8 m/s2 Often, y0 = 0. Sometimes v0 = 0**Equations for Objects in Free Fall**• Written taking “down” as + y v = v0 + gt (1) y = y0 + v0t + (½)gt2 (2) (v)2 = (v0)2 + 2g(y - y0) (3) vavg = (½)(v + v0) (4) g = 9.8 m/s2 Often, y0 = 0. Sometimes v0 = 0**ExampleFalling from a tower (v0 = 0)**Note! Take y as positive DOWNWARD! v = at y = (½)at2 a = g = 9.8 m/s2**ExampleFalling from a tower (v0 = 0)**Note! Take y as positive DOWNWARD! v = at y = (½)at2 a = g = 9.8 m/s2**ExampleFalling from a tower (v0 = 0)**Note! Take y as positive DOWNWARD! v = at y = (½)at2 a = g = 9.8 m/s2**ExampleFalling from a tower (v0 = 0)**v1 = (9.8)(1) = 9.8 m/s Note! Take y as positive DOWNWARD! v = at y = (½)at2 a = g = 9.8 m/s2 v2 = (9.8)(2) = 19.6 m/s v3 = (9.8)(3) = 29.4 m/s**Example:**Thrown Down From a Tower A ball is thrown downwardwith initial velocity v0 = 3.0 m/s, instead of being dropped. (a)Position after t = 1.0 s & 2.0 s? y = v0t + (½)at2 t = 1.0 s; y = (3)(1) + (½)(9.8)(1)2 = 7.9m t = 2.0s ; y = (3)(2) + (½)(9.8)(2)2 = 25.6 m Photo of the leaning tower of Pisa**Example:**Thrown Down From a Tower A ball is thrown downwardwith initial velocity v0 = 3.0 m/s, instead of being dropped. (a)Position after t = 1.0 s & 2.0 s? y = v0t + (½)at2 t = 1.0 s; y = (3)(1) + (½)(9.8)(1)2 = 7.9m t = 2.0s ; y = (3)(2) + (½)(9.8)(2)2 = 25.6 m Photo of the leaning tower of Pisa**Example:**Thrown Down From a Tower A ball is thrown downwardwith initial velocity v0 = 3.0 m/s, instead of being dropped. (a)Position after t = 1.0 s & 2.0 s? y = v0t + (½)at2 t = 1.0 s; y = (3)(1) + (½)(9.8)(1)2 = 7.9m t = 2.0s ; y = (3)(2) + (½)(9.8)(2)2 = 25.6 m (b) Speed after t =1.0 s & 2.0 s? v = v0 + at t = 1.0 s; v = 3 + (9.8)(1) = 12.8m/s t = 2.0s ; v = 3 + (9.8)(2) = 22.6m/sCompare with speeds of a dropped ball. Photo of the leaning tower of Pisa**A Useful, Detailed Example**v = 0 here but still have a = -g A person throws a ball up into the air with initial velocity v0 = 15.0 m/s. Questions: a. Time to the top? b. Round trip time? c. Maximum height? d. Velocity when it comes back to the start? e. Times when the height y = 8.0 m? Time to the top = (½) round trip time! vC = -vA (= -v0) Work in general on the white board! vA = v0 = 15 m/s choose y as positive upward a = -g = - 9.8 m/s2**Example: Ball thrown upward; the quadratic formula. For a**ball thrown upward at an initial speed of v0 = 15.0 m/s, calculate the times t the ball passes a point y = 8.0 m above the person’s hand.**Example: A ball thrown upward at the edge of a cliff.**A ball is thrown up with initial velocity v0 = 15.0 m/s, by a person standing on the edge of a cliff, so that it can fall to the base of the cliff 50.0 m below. Calculate: a. The time it takes the ball to reach the base of the cliff. b. The total distance traveled by the ball.**Example: Not a bad throw for a rookie!**A stone is thrown at point (A) from the top of a building with initial velocity v0 = 19.2 m/s straight up. The building is H = 49.8 m high, & the stone just misses the edge of the roof on its way down, as in the figure. Calculate: a) The time at which it reaches its maximum height.b) It’s maximum height above the rooftop.c) The time at which it returns to the thrower’s hand. d) It’s velocity when it returns to the thrower’s hand. e) It’s velocity & position at time t = 5 s.