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A ball is dropped...

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|     o       Location 1 (175 cm up in the air)
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|     o       Location 2
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|_____o______ Location 3 (ground)
  
The "o" is supposed to represent a falling ball with a mass of 100 grams. It is dropped from a height of 175 cm. The ball has a velocity when it hits the ground. What is the height of its location when its velocity is half of that (location 2)? Okay, at location 1, the ball has a velocity of 0, and hence, has no kinematic energy. It has a potential energy of mgh though. At location 3, it has a velocity of v. Its kinematic energy is mv^2 / 2. It has no potential energy. The energy principle gives us: mv^2 / 2 = mgh v^2 / 2 = gh v^2 = 2gh v = sqrt(2gh) At location 2, the ball has a velocity of v / 2. Its kinematic energy is m(v/2)^2 / 2, and its potential energy is mgh. Our beloved energy principle gives us: m(v/2)^2 / 2 + mgh = mv^2 / 2 m(v/2)^2 + 2mgh = mv^2 mv^2/4 + 2mgh = mv^2 mv^2 + 8mgh = 4mv^2 8mgh = 4mv^2 - mv^2 8gh = 3v^2 h = 3v^2/8g Plugging in v = sqrt(2gh) gives us a totally wrong answer. The answer is supposed to be 175/4 cm. What am I doing wrong? [edited by - Muzzafarath on January 31, 2003 3:05:22 PM]

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Guest Anonymous Poster
Doing this numerically seems to work.

At top: potential energy (PE) = mgh = 0.1 x 9.8 x 1.75 = 1.715J
At bottom: PE = kinetic energy (KE) so KE = 0.5 x m x v^2 = 1.715J
v = sqrt((1.715 x 2) / 0.1) = sqrt(34.3)

At middle: v = sqrt(34.3) / 2 so KE = 0.5 x 0.1 x (sqrt(34.3) / 2)^2 = 0.42875J

PE = 1.715 - 0.42875 = 1.28625J since PE + KE = 1.715 (max E)
mgh = 1.28625J so h = 1.28625 / (0.1 x 9.8) = 1.3125m = (1.75 x 3) / 4 m = (175 x 3) / 4 cm.

Almost right.

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Doh, I interpreted the question wrong. The ball is indeed (175 * 3) / 4 = 131,25 cm from the ground (as AP said), which is in fact the same place as 175 - 131,25 = 43,75 = 175/4 cm from "dropping position". *bangs head against desk*

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You can also resolve your problem using projectils (bad spelling i know)


y = y0 + v0yt+ 1/2at^2

were y is the position
y0 is the inicial position
t is time (duh)
a is the aceleration in this case gravity
and v0 is the inicial position

you can use this to find

and to find the velocity you can use

v = v0y + at

hope it helps in any way


[edited by - force_of_will on January 31, 2003 5:20:36 PM]

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