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1. Identify the knowns. y 0 = 0 ; y 1 = 5 . 10 m ; v 0 = 13 .0 m/s ; a = g = 9 . 80 m /s 2 size 12{a= - g= - 9 "." "80"" m/s" rSup { size 8{2} } } {} .

2. Choose the kinematic equation that makes it easiest to solve the problem. The equation v 2 = v 0 2 + 2 a ( y y 0 ) works well because the only unknown in it is v . (We will plug y 1 in for y .)

3. Enter the known values

v 2 = 13 . 0 m/s 2 + 2 9 . 80 m/s 2 5 . 10 m 0 m = 268 . 96 m 2 /s 2 , size 12{v rSup { size 8{2} } = left ( - "13" "." "0 m/s" right ) rSup { size 8{2} } +2 left ( - 9 "." "80 m/s" rSup { size 8{2} } right ) left ( - 5 "." "10 m" - "0 m" right )="268" "." "96 m" rSup { size 8{2} } "/s" rSup { size 8{2} } } {}

where we have retained extra significant figures because this is an intermediate result.

Taking the square root, and noting that a square root can be positive or negative, gives

v = ± 16 .4 m/s .

The negative root is chosen to indicate that the rock is still heading down. Thus,

v = 16 .4 m/s . size 12{v= - "16" "." 4`"m/s"} {}


Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed . (See [link] and [link] (a).) This is not a coincidental result. Because we only consider the acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical position relative to the starting point. For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from [link] ) when the initial velocity is 13.0 m/s straight up, a result of ± 3 . 20 m/s size 12{ +- 3 "." "20"`"m/s"} {} is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction.

Two figures are shown. At left, a man standing on the edge of a cliff throws a rock straight up with an initial speed of thirteen meters per second. At right, the man throws the rock straight down with a speed of thirteen meters per second. In both figures, a line indicates the rock’s trajectory. When the rock is thrown straight up, it has a speed of minus sixteen point four meters per second at minus five point one zero meters below the point where the man released the rock. When the rock is thrown straight down, the velocity is the same at this position.
(a) A person throws a rock straight up, as explored in [link] . The arrows are velocity vectors at 0, 1.00, 2.00, and 3.00 s. (b) A person throws a rock straight down from a cliff with the same initial speed as before, as in [link] . Note that at the same distance below the point of release, the rock has the same velocity in both cases.

Another way to look at it is this: In [link] , the rock is thrown up with an initial velocity of 13 .0 m/s . It rises and then falls back down. When its position is y = 0 on its way back down, its velocity is 13 .0 m/s . That is, it has the same speed on its way down as on its way up. We would then expect its velocity at a position of y = 5 . 10 m to be the same whether we have thrown it upwards at + 13 .0 m/s or thrown it downwards at 13 .0 m/s . The velocity of the rock on its way down from y = 0 is the same whether we have thrown it up or down to start with, as long as the speed with which it was initially thrown is the same.

Find g From data on a falling object

The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics laboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, [link] . Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.

Questions & Answers

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Mahmud Reply
The Critical Angle Derivation So the critical angle is defined as the angle of incidence that provides an angle of refraction of 90-degrees. Make particular note that the critical angle is an angle of incidence value. For the water-air boundary, the critical angle is 48.6-degrees.
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the ratio of the density of a substance to the density of a standard, usually water for a liquid or solid, and air for a gas.
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Sean Reply
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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