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Student determine the effect of gravity on objects without support. Students will calculate these effects of gravity over time. 1 Free Fall

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V f = V i + at d = V i t + ½ at 2 2ad = V f 2 – V i 2 V f + V i 2 2 Four Basic Equations of Physics d =t

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Free fall – motion under the influence of the gravitational force only (neglects air resistance) 3 Free Fall

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time that has passed from the beginning of a fall 4 Elapsed time

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Acceleration due to gravity is 9.8 m/s 2, downward Every second that an object falls, the velocity increases by 9.8 m/s 5 Gravity and Free fall

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If an object is dropped from rest at the top of a cliff, how fast will it be going after 1 second? 2 seconds? 10 seconds? 6 Sample 1

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When an object is falling, assume that a = 9.8 m/s 2 “from rest” tells us that V i = 0 “how fast will it be going?” is asking us to find V f = ? (this is the unknown) the first problem gives us a time of t = 1 s 7 First, Identify the GIVENS?

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G: a = 9.8 m/s2 V i = 0 t = 1 s U: V f = ? E: V f = V i + at S: V f = 0 + (9.8 m/s 2) (1 s) S: V f = 9.8 m/s 8 Therefore…

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G: a = 9.8 m/s 2 V i = 0 t = 2 s U: V f = ? E: V f = V i + at S: V f = 0 + (9.8 m/s 2) (2 s) S: V f = 19.6 m/s 9 Now find out how fast it will be falling after 2 s…

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G: a = 9.8 m/s 2 V i = 0 t = 10 s U: V f = ? E: V f = V i + at S: V f = 0 + (9.8 m/s 2) (10 s) S: V f = 98 m/s 10 Now find out how fast it will be falling after 10 s…

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If we use V f = V i + at, and V i = 0 then we actually have an equation that reads… V f = at, where a = 9.8 m/s 2. When a = 9.8 m/s 2, we call that constant g. Therefore, we can use the following new, or derived equation… Instantaneous speed = acceleration X elapsed time v = gt 11 Calculating How Fast Free Fall Is

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The Demon Drop ride at Cedar Point Amusement Park falls freely for 1.5 s after starting from rest. What is its velocity at the end of 1.5 s? How far does it fall? 12 Sample Problem #2

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G: a = 9.8 m/s 2 V i = 0 (starting from rest) t = 1.5 s U: V f = ? E: V f = V i + at S: V f = 0 + (9.8 m/s 2) (1.5 s) S: V f = 14.7 m/s 13 What is its velocity at the end of 1.5 s?

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G: a = 9.8 m/s 2 V i = 0 t = 1.5 s V f = 14.7 m/s (from previous part) U: d = ? (how far) E: d = V i t + ½ at 2 (use any of the 3 motion formulas with d in them) S: d = (0)(1.5s) + ½ (9.8 m/s 2 )(1.5s) 2 S: d = 11.03 m 14 How far does it fall?

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If we use d = V i t + ½ at 2, and V i = 0 then we actually have an equation that reads… d = ½ at 2, where a = 9.8 m/s 2. When a = 9.8 m/s 2, we call that constant g. Therefore, we can use the following new, or derived equation… d = ½ gt 2 15 Calculating How Far Free Fall Is

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If the object is moving down, a = 9.8 m/s 2 If the object is moving up, a = - 9.8 m/s 2 16 Acceleration due to Gravity a = 9.8 m/s 2 (speeding up) a = - 9.8 m/s 2 (slowing down)

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When an object is thrown into the air, the velocity at its highest point is ZERO!!! 17 Throwing an object into the air Begin V i = ? Begin V i = 0 m/s End V f = 0 m/s End V f = ?

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A ball is thrown vertically into the air with an initial velocity of 4 m/s. How high does the ball rise? How long does it take to reach its highest point? If the ball is caught in the same spot from which it was thrown, what is the total amount of time that it was in the air? What is its velocity just before it is caught? 18 Sample Problem #3

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HINT: draw a picture and label it 19 How high does the ball rise? Begin V i = 4 m/s End V f = 0 m/s d= ?

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G: a = -9.8 m/s 2 (notice the negative sign, ball moving upward) V i = 4 m/s V f = 0 m/s (at the top, before it starts to fall, it stops) U: d = ? E: 2ad = V f 2 – V i 2, (solve for d) d = V f 2 – V i 2 2a S: d = 0 2 – (4 m/s) 2 2(-9.8 m/s 2 ) S: d = 0.816 m 20 How high does the ball rise?

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G: a = -9.8 m/s 2 U: t = ? V i = 4 m/s V f = 0 m/s d = 0.816 m E: V f = V i + at, (solve for t) t = V f – V i a S: t = 0 m/s – 4 m/s ( -9.8 m/s 2 ) S: t = 0.408 s 21 How long does it take to reach its highest point?

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G: a = 9.8 m/s 2 (ball going down, positive) U: t = ? V i = 4 m/s V f = 0 m/s d = 0.816 m E: d = V i t + ½ at 2 ; derived to d = ½ gt 2, solve for t… t 2 = (2d)/g S: t 2 = (2)(0.816 m)/(9.8 m/s 2 ) *don’t forget to take the square root S: t = 0.408s Total time = time Going up + time going down Total time = 0.408s +0.408s = 0.816s 22 If the ball is caught in the same spot from which it was thrown, what is the total amount of time that it was in the air?

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G: a = 9.8 m/s 2 (ball moving down before it is caught) V i = 0 m/s d = 0.816 m t = 0.408 s U: V f = ? E: 2ad = V f 2 – V i 2, (solve for V f ) V f 2 = 2ad + V i 2 S: V f 2 = 2(9.8 m/s 2 )(0.816 m) + 0 2 S: V f = 3.99 or 4 m/s 23 What is its velocity just before it is caught?

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You’re done! Now try some problems on your own. 24

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Keep your projects at your seat for now. Place your completed Lab in the TOP tray. Someone pass out all papers in the 4A drawer. Complete the warm-ups for today.

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