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Falling Objects

Falling Objects. Freely Falling Objects. An important & common special case of uniformly accelerated motion is “ FREE FALL ”. Objects falling in Earth’s gravity. Neglect air resistance. Use the one dimensional uniform acceleration equations (with some changes in notation, as we will see).

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Falling Objects

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  1. Falling Objects

  2. Freely Falling Objects • An important & common special case of uniformly accelerated motion is “FREE FALL” • Objects falling in Earth’s gravity. Neglect air resistance. Use the one dimensional uniform acceleration equations (with some changes in notation, as we will see)

  3. Falling Objects Experiment • A ball & a light piece of paper dropped at the same time. Repeated with wadded up paper.

  4. The Acceleration due to • gravity at 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 • A rock & a feather are dropped at the same time in air. Repeated in vacuum.

  5. 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 is: g = 9.8 m/s2 (approximately!)

  6. 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.

  7. Near Earth’s surface, all objects experience • approximatelythe 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.

  8. Falling Objects

  9. Acceleration Due to Gravity g = 9.8 m/s2(approximately) • Depends on location on Earth, latitude, & altitude:

  10. Note: My treatment is slightly different than the book’s, but it is, of course, equivalent! • To treat falling objects, use the same equations we already have, but change notation slightly: Replace a by g = 9.8 m/s2 But in the equations it could have a + or a – sign in front of it! 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)

  11. NOTE!!! • Whenever I (or the author!) write the symbol g, it ALWAYS means the POSITIVE numerical value 9.8 m/s2! It NEVER is negative!!! • The sign (+ or -) of the gravitational acceleration is taken into account in the equations we now discuss!

  12. The Sign of g in 1d Equations The magnitude (size) of g = 9.8 m/s2 (Always a Positive Number!) • But, acceleration is avector (1 dimen), with 2 possible directions. Call these + and -. • However, which way is + which way is – is ARBITRARY & UP TO US! • May seem “natural” for “up” to be + y and “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!

  13. Directions of Velocity & Acceleration • Objects in free fall ALWAYShave downward acceleration. • Still use the same equations for objects thrown upward with some initial velocity v0 • An object goes up until it stops at some point & then it falls back down. Acceleration is alwaysgin the downward direction. For the first half of flight, the velocity is UPWARD.  For the first part of the flight, velocity & acceleration are in opposite directions!

  14. Equations for Objects in Free Fall Written taking “up” as+ y! v = v0 - gt (1) y = y0 + v0 t – (½)gt2 (2) v2 = (v0)2 - 2g (y - y0) (3) vave = (½)(v + v0) (4) g = 9.8 m/s2 Usually y0 = 0. Sometimes v0 = 0

  15. Equations for Objects in Free Fall Written taking “down” as+ y! v = v0 + gt (1) y = y0 + v0 t + (½)gt2 (2) v2 = (v0)2 + 2g (y - y0) (3) vave = (½)(v + v0) (4) g = 9.8 m/s2 Usually y0 = 0. Sometimes v0 = 0

  16. Example: Falling from a Tower A ball is dropped (v0 = 0) from a tower 70.0 m high. How far will it have fallen after time t1 = 1 s, t2 = 2 s, t3 = 3 s? Note:y is positive DOWNWARD! v = gt, y = (½) gt2 a = g = 9.8 m/s2 v1 = (9.8)(1) = 9.8 m/s v3 = (9.8)(3) = 29.4 m/s v2 = (9.8)(2) = 19.6 m/s

  17. Example: Thrown Down From a Tower A ball is thrown down with an initial downward velocity of v0 = 3 m/s, instead of being dropped. What are it’s position & speed aftert1 = 1 s,t2 = 2 s, t3 = 3 s?Compare with the dropped ball. y is positive DOWNWARD! v = v0 + gt y = v0t + (½)gt2 a = g = 9.8 m/s2 v1 = 3 + (9.8)(1) = 12.8 m/s y1 = (3)(1) + (½)(9.8)(1)2 = 3.0 + 4.9 = 7.9 m v2 = 3 + (9.8)(2) = 20.6 m/s y2 = (3)(2) + (½)(9.8)(2)2 = 6.0 + 19.6 = 25.6 m v3 = 3 + (9.8)(3) = 31.5 m/s y3 = (3)(3) + (½)(9.8)(3)2 = 9.0 + 44.1 = 53.1 m

  18. Example: • A person throws a ball up into the air • with an initial velocity v0 = 15.0 m/s. • Calculate: • a.The time to reach maximum height. • b.The maximum height. • c.The time to come back to the hand. • d. Velocity when it returns to the hand. • Note:y is positiveUPWARD! • v = v0 – gt, y = v0t - (½)gt2 • v2 = (v0)2 - 2g(y - y0) • Work this in general on the • white board! v= 0 here, but a = - g! Time to top = ½ round trip time! vA = 15 m/s  vC = - v0 = -15 m/s

  19. Example: A ball is thrown upward at an initial • speed v0 = 15.0 m/s, calculate the timest the ball • passes a point y = 8.0 mabove the person’s hand.

  20. Example:A ball thrown upward at the edge of a cliff. A ball is thrown upward 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. a. Use y = v0t – (½)gt2 -50 = (15)t – (½)(9.8)t2(quadratic equation!) Solutions: t = 5.07 s, t = -2.01 s What is the meaning of t = -2.01 s? It has no physical meaning! It is absurd!

  21. Example:A ball thrown upward at the edge of a cliff. A ball is thrown upward 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. b. ymax = (v0)2/(2g) = 11.5 m Distance Traveled = 2(ymax) + 50 = 2(11.5) + 50 = 73 m

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