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2.6 Falling objects  (Page 5/9)

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Figure has four panels. The first panel (on the top) is an illustration of a ball falling toward the ground at intervals of one tenth of a second. The space between the vertical position of the ball at one time step and the next increases with each time step. At time equals 0, position and velocity are also 0. At time equals 0 point 1 seconds, y position equals negative 0 point 049 meters and velocity is negative 0 point 98 meters per second. At 0 point 5 seconds, y position is negative 1 point 225 meters and velocity is negative 4 point 90 meters per second. The second panel (in the middle) is a line graph of position in meters versus time in seconds. Line begins at the origin and slopes down with increasingly negative slope. The third panel (bottom left) is a line graph of velocity in meters per second versus time in seconds. Line is straight, beginning at the origin and with a constant negative slope. The fourth panel (bottom right) is a line graph of acceleration in meters per second squared versus time in seconds. Line is flat, at a constant y value of negative 9 point 80 meters per second squared.
Positions and velocities of a metal ball released from rest when air resistance is negligible. Velocity is seen to increase linearly with time while displacement increases with time squared. Acceleration is a constant and is equal to gravitational acceleration.

Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location?

Strategy

Draw a sketch.

The figure shows a green dot labeled v sub zero equals zero meters per second, a purple downward pointing arrow labeled a equals question mark, and an x y coordinate system with the y axis pointing vertically up and the x axis pointing horizontally to the right.

We need to solve for acceleration a size 12{a} {} . Note that in this case, displacement is downward and therefore negative, as is acceleration.

Solution

1. Identify the knowns. y 0 = 0 ; y = –1 .0000 m ; t = 0 .45173 ; v 0 = 0 size 12{v rSub { size 8{0} } =0} {} .

2. Choose the equation that allows you to solve for a size 12{a} {} using the known values.

y = y 0 + v 0 t + 1 2 at 2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}

3. Substitute 0 for v 0 size 12{v rSub { size 8{0} } } {} and rearrange the equation to solve for a size 12{a} {} . Substituting 0 for v 0 size 12{v rSub { size 8{0} } } {} yields

y = y 0 + 1 2 at 2 . size 12{y=y rSub { size 8{0} } + { {1} over {2} } ital "at" rSup { size 8{2} } "." } {}

Solving for a size 12{a} {} gives

a = 2 y − y 0 t 2 . size 12{a= { {2 left (y - y rSub { size 8{0} } right )} over {t rSup { size 8{2} } } } "." } {}

4. Substitute known values yields

a = 2 ( − 1 . 0000 m – 0 ) ( 0 . 45173 s ) 2 = − 9 . 8010 m/s 2 , size 12{a= { {2 \( - 1 "." "0000 m–0" \) } over { \( 0 "." "45173 s" \) rSup { size 8{2} } } } = - 9 "." "8010 m/s" rSup { size 8{2} } ,} {}

so, because a = − g size 12{a= - g} {} with the directions we have chosen,

g = 9 . 8010 m/s 2 . size 12{g=9 "." "8010 m/s" rSup { size 8{2} } } {}

Discussion

The negative value for a size 12{a} {} indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of 9 . 80 m/s 2 size 12{9 "." "80 m/s" rSup { size 8{2} } } {} , so 9 . 8010 m/s 2 size 12{9 "." "8010 m/s" rSup { size 8{2} } } {} makes sense. Since the data going into the calculation are relatively precise, this value for g size 12{g} {} is more precise than the average value of 9 . 80 m/s 2 size 12{9 "." "80 m/s" rSup { size 8{2} } } {} ; it represents the local value for the acceleration due to gravity.

A chunk of ice breaks off a glacier and falls 30.0 meters before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water?

We know that initial position y 0 = 0 , final position y = −30 . 0 m , and a = − g = − 9 . 80 m/s 2 . We can then use the equation y = y 0 + v 0 t + 1 2 at 2 to solve for t . Inserting a = − g , we obtain

y = 0 + 0 − 1 2 gt 2 t 2 = 2 y − g t = ± 2 y − g = ± 2 ( − 30.0 m ) − 9.80 m /s 2 = ± 6.12 s 2 = 2.47 s ≈ 2.5 s

where we take the positive value as the physically relevant answer. Thus, it takes about 2.5 seconds for the piece of ice to hit the water.

Phet explorations: equation grapher

Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. y = bx size 12{y= ital "bx"} {} ) to see how they add to generate the polynomial curve.

Equation Grapher

Section summary

  • An object in free-fall experiences constant acceleration if air resistance is negligible.
  • On Earth, all free-falling objects have an acceleration due to gravity g size 12{g} {} , which averages
    g = 9 . 80 m/s 2 . size 12{g=9 "." "80 m/s" rSup { size 8{2} } } {}
  • Whether the acceleration a should be taken as + g size 12{+g} {} or − g is determined by your choice of coordinate system. If you choose the upward direction as positive, a = − g = − 9 . 80 m /s 2 is negative. In the opposite case, a = +g = 9 . 80 m/s 2 is positive. Since acceleration is constant, the kinematic equations above can be applied with the appropriate + g or − g substituted for a .
  • For objects in free-fall, up is normally taken as positive for displacement, velocity, and acceleration.

Conceptual questions

What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down?
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Source:  OpenStax, Cc test coll. OpenStax CNX. Dec 15, 2015 Download for free at http://legacy.cnx.org/content/col11717/1.4
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