## Student determine the effect of gravity on objects without support. ## Student determine the effect of gravity on objects without support. ## Students will calculate these effects of gravity over time.
**Vf = Vi + at** **Vf = Vi + at**
**d = Vi t + ½ at2**
**2ad = Vf2 – Vi2**
** Vf + Vi **
** 2**
## Free fall – motion under the influence of the gravitational force only (neglects air resistance) ## Free fall – motion under the influence of the gravitational force only (neglects air resistance)
## Acceleration due to gravity is 9.8 m/s2, downward ## Acceleration due to gravity is 9.8 m/s2, downward ## Every second that an object falls, the velocity increases by 9.8 m/s
## If an object is dropped from rest at the top of a cliff, how fast will it be going after 1 second? ## 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?
## 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
## G: a = 9.8 m/s2 ## G: a = 9.8 m/s2 ## U: Vf = ? ## E: Vf = Vi + at ## S: Vf = 0 + (9.8 m/s2) (1 s) ## S: Vf = 9.8 m/s
## G: a = 9.8 m/s2 ## 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
## G: a = 9.8 m/s2 ## 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
## 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… - 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…
## The Demon Drop ride at Cedar Point Amusement Park falls freely for 1.5 s after starting from rest. ## 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?
## G: a = 9.8 m/s2 ## 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
## G: a = 9.8 m/s2 ## 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
## If we use d = Vit + ½ at2 , and Vi = 0 then we actually have an equation that reads… - 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
## If the object is moving down, a = 9.8 m/s2 ## If the object is moving up, a = - 9.8 m/s2
## When an object is thrown into the air, the velocity at its highest point is __ZERO__!!!
## A ball is thrown vertically into the air with an initial velocity of 4 m/s. ## 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?
## HINT: draw a picture and label it
## G: a = -9.8 m/s2 (notice the negative sign, ball moving upward) ## 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
## G: a = -9.8 m/s2 U: t = ? - 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
## G: a = 9.8 m/s2 (ball going down, positive) U: t = ? - 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
## G: a = 9.8 m/s2 (ball moving down 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! ## You’re done! ## Now try some problems on your own.
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