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

 Sana 05.03.2018 Hajmi 447 b.

• Vi = 0
• t = 1 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

• ## 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…

• v = gt

• ## 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

• 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…

• ## 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?

• ## 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)

• 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

• ## Now try some problems on your own.

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