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

Four Basic Equations of Physics Vf = Vi + at d = Vi t + ½ at2 2ad = Vf2 – Vi2 Vf + Vi 2 d = t

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

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

Gravity and Free fall Acceleration due to gravity is 9.8 m/s2, downward Every second that an object falls, the velocity increases by 9.8 m/s

Sample 1 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?

First, Identify the GIVENS? When an object is falling, assume that a = 9.8 m/s2 “from rest” tells us that Vi = 0 “how fast will it be going?” is asking us to find Vf = ? (this is the unknown) the first problem gives us a time of t = 1 s

Therefore… G: a = 9.8 m/s2 U: Vf = ? E: Vf = Vi + at t = 1 s U: Vf = ? E: Vf = Vi + at S: Vf = 0 + (9.8 m/s2) (1 s) S: Vf = 9.8 m/s

Now find out how fast it will be falling after 2 s… G: a = 9.8 m/s2 Vi = 0 t = 2 s U: Vf = ? E: Vf = Vi + at S: Vf = 0 + (9.8 m/s2) (2 s) S: Vf = 19.6 m/s

Now find out how fast it will be falling after 10 s… G: a = 9.8 m/s2 Vi = 0 t = 10 s U: Vf = ? E: Vf = Vi + at S: Vf = 0 + (9.8 m/s2) (10 s) S: Vf = 98 m/s

Calculating How Fast Free Fall Is If we use Vf = Vi + at, and Vi = 0 then we actually have an equation that reads… Vf = at, where a = 9.8 m/s2. When a = 9.8 m/s2, we call that constant g. Therefore, we can use the following new, or derived equation… Instantaneous speed = acceleration X elapsed time v = gt

Sample Problem #2 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?

What is its velocity at the end of 1.5 s? G: a = 9.8 m/s2 Vi = 0 (starting from rest) t = 1.5 s U: Vf = ? E: Vf = Vi + at S: Vf = 0 + (9.8 m/s2) (1.5 s) S: Vf = 14.7 m/s

How far does it fall? Vi = 0 t = 1.5 s G: a = 9.8 m/s2 Vi = 0 t = 1.5 s Vf = 14.7 m/s (from previous part) U: d = ? (how far) E: d = Vit + ½ at2 (use any of the 3 motion formulas with d in them) S: d = (0)(1.5s) + ½ (9.8 m/s2 )(1.5s)2 S: d = 11.03 m

Calculating How Far Free Fall Is If we use d = Vit + ½ at2 , and Vi = 0 then we actually have an equation that reads… d = ½ at2 , where a = 9.8 m/s2. When a = 9.8 m/s2, we call that constant g. Therefore, we can use the following new, or derived equation… d = ½ gt2

Acceleration due to Gravity If the object is moving down, a = 9.8 m/s2 If the object is moving up, a = - 9.8 m/s2 a = 9.8 m/s2 (speeding up) a = - 9.8 m/s2 (slowing down)

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

Sample Problem #3 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?

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

How high does the ball rise? G: a = -9.8 m/s2 (notice the negative sign, ball moving upward) Vi = 4 m/s Vf = 0 m/s (at the top, before it starts to fall, it stops) U: d = ? E: 2ad = Vf2 – Vi2, (solve for d) d = Vf2 – Vi2 2a S: d = 02 – (4 m/s)2 2(-9.8 m/s2 ) S: d = 0.816 m

How long does it take to reach its highest point? G: a = -9.8 m/s2 U: t = ? Vi = 4 m/s Vf = 0 m/s d = 0.816 m E: Vf = Vi + at, (solve for t) t = Vf – Vi a S: t = 0 m/s – 4 m/s (-9.8 m/s2) S: t = 0.408 s

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? G: a = 9.8 m/s2 (ball going down, positive) U: t = ? Vi = 4 m/s Vf = 0 m/s d = 0.816 m E: d = Vit + ½ at2 ; derived to d = ½ gt2, solve for t… t2 = (2d)/g S: t2 = (2)(0.816 m)/(9.8 m/s2) *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

What is its velocity just before it is caught? G: a = 9.8 m/s2 (ball moving down before it is caught) Vi = 0 m/s d = 0.816 m t = 0.408 s U: Vf = ? E: 2ad = Vf2 – Vi2, (solve for Vf) Vf2 = 2ad + Vi2 S: Vf2 = 2(9.8 m/s2 )(0.816 m) + 02 S: Vf = 3.99 or 4 m/s

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

Keep your projects at your seat for now 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.