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Wize High School Grade 12 Physics Textbook > Kinematics

Free Fall

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Free Fall


  • Near the Earth's surface, all objects experience the same acceleration due to gravity (neglecting air resistance). This acceleration has magnitude g=9.81m/s2g = 9.81 m/s^2 and points downward, to the center of the Earth.

  • Note that this value changes if you are extremely far away from Earth's surface, or if you are on another planet (or moon)!

  • To solve problems with freely falling objects, we simply use g in the kinematic equations in place of the acceleration, a. If you define the downwards direction to be negative, then you will use -g. Even if the object is thrown upward, the gravitational acceleration still points downward.
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Example: Free Fall


I drop a ball off of a 100m tall building.

a) What is its velocity just before it hits the ground?
b) What is its average velocity?
c) What is the instantaneous velocity when the ball is halfway down the building?
Use g=9.81 m/s2.

Solutions:


v0=0msvf=?d=100 ma=9.81 m/s2\begin{array}{l} v_0=0\frac{m}{s}\\ v_f=?\\ d=100~m\\ a=9.81~m/s^2 \end{array}

Part a)


I choose downward direction to be positive. So, any vector pointing downward is positive and any vector pointing upward is negative. (Alternatively you can choose upward direction to be positive. As long as you remain consistent, you should get the same answer at the end)

Use the full distance, 100 meters.
vf2=v02+2advf=v02+2ad=02+2(9.81m/s2)(100m)=44.3 m/s before it hits the ground\begin{array}{lll} v_f^2=v_0^2+2ad\\ v_f=\sqrt{v_0^2+2ad}\\[3pt] =\sqrt{0^2+2(9.81 m/s^2)(100m)}\\[2pt] =44.3~m/s~\textrm{before it hits the ground} \end{array}

Part b)


When acceleration is constant, the average velocity can be determined simply by averaging the initial and final velocities:

vavg=v1 + v22=0 + 44.32=22.1 m/sv_{avg}=\frac{v_{1}\ +\ v_{2}}{2}=\frac{0\ +\ 44.3}{2}=22.1~m/s

Note: we are defining the downward direction as positive, so we leave all answers as positive. If you picked the frame of reference pointing upward, your answers would need to be answered as negative. (If instead we asked for SPEED instead of velocity, then you would enter all positive values because speed is a scalar.)



Part c)


Use half of the distance (50 meters).
vf2=v02+2adv0=0msvf=v02+2advf=?=02+2(9.81m/s2)(50m)d=100 m=31.3 m/s halfway down the buildinga=9.81 m/s2\begin{array}{lll} v_f^2=v_0^2+2ad&&v_0=0\frac{m}{s}\\ v_f=\sqrt{v_0^2+2ad}&&v_f=?\\[3pt] =\sqrt{0^2+2(9.81m/s^2)(50m)}& &d=100~m\\[3pt] =31.3~m/s~\textrm{halfway down the building} && a=9.81~m/s^2 \end{array}




A boy decided to estimate the height of a two-storey building by throwing a stone up and measuring the time it takes for the stone to reach the top of the building. Assume he throws the stone from 1 meter above the ground, such that the stone stops and turns back just when it reached the top of the building.

If the stone takes 2s to reach the top of the building, what is the initial velocity of the stone? (Pick upward direction to be positive)