Student determine the effect of gravity on objects without support. Student determine the effect of gravity on objects without support


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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

  • 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

      • Vi = 0
      • t = 1 s
  • 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…
  • Instantaneous speed = acceleration X elapsed time

    • v = gt


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 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!!!

  • 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

  • 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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