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)

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

Experiment A rock & a feather are dropped at the same time in air. Repeated in vacuum. 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.

g = 9.8 m/s2 The magnitude of the acceleration due to gravity is: 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!)

Acceleration due to gravity, g = 9.8 m/s2 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.

resistance, all objects Near Earth’s surface, 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. Figure 2-26. Caption: Multiflash photograph of a falling apple, at equal time intervals. The apple falls farther during each successive interval, which means it is accelerating.

Falling Objects

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

But in the equations it could have 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)

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!

The Sign of g in 1d Equations The magnitude (size) of g = 9.8 m/s2 (Always a Positive Number!) But, acceleration is a vector (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!

Directions of Velocity & Acceleration Objects in free fall ALWAYS have 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 always g in 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!

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

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

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

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 after t1 = 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

Work this in general on the 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 positive UPWARD! 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

Example: A ball is thrown upward at an initial speed v0 = 15.0 m/s, calculate the times t the ball passes a point y = 8.0 m above the person’s hand. Figure 2-31. Caption: Graphs of (a) y vs. t, (b) v vs. t for a ball thrown upward, Examples 2–16, 2–18, and 2–19. Solution: We are given the initial and final position, the initial speed, and the acceleration, and want to find the time. This is a quadratic equation; there are two solutions: t = 0.69 s and t = 2.37 s. The first is the ball going up and the second is the ball coming back down.

Calculate: 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! Figure 2-32. Caption: Example 2–20. The person in Fig. 2–30 stands on the edge of a cliff. The ball falls to the base of the cliff, 50.0 m below. Solution: a. We use the same quadratic formula as before, we find t = 5.07 s (the negative solution is physically meaningless). b. The ball goes up 11.5 m, then down 11.5 m + 50 m, for a total distance of 73.0 m.

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 Figure 2-32. Caption: Example 2–20. The person in Fig. 2–30 stands on the edge of a cliff. The ball falls to the base of the cliff, 50.0 m below. Solution: a. We use the same quadratic formula as before, we find t = 5.07 s (the negative solution is physically meaningless). b. The ball goes up 11.5 m, then down 11.5 m + 50 m, for a total distance of 73.0 m.