Question

As astronaut visiting Planet X, you’re assigned to measure the free-fall acceleration. Getting out your meter stick and stopwatch, you time the fall of a heavy ball from several Heights. You data is as follows:
Height (m) Fall Time (s)
0.0 0.00
1.0 0.54
2.0 0.72
3.0 0.91
4.0 1.01
5.0 1.17
a.) Analyze these data to determine the free-fall acceleration on Planet X. Your analysis should involve fitting a straight lie to an appropriate graph. Place the graph here. (Hint Kinematics)
b.) Determine the uncertainty in the free-fall acceleration.

Answer 1

Answer:

g = (0.13 + - 0.05) m / s²

Explanation:

a) The equations of physics are valid in all inertial systems, therefore the kinematic relationships on this planet are the same as on Earth.

        y = v₀ t - ½ g t²

From the statement it follows that the body is released, so its initial velocity is zero

        y = 1/2 g t²

where we take the downward direction as positive

With this expression to make a linear graph we must graph t² vs y, see attached.  Observe that the controlled variable is the height, so this is the independent variable and must go on the x axis, the values ​​of the slope and point of cut can be obtained from this graph  by lineal regretion

           y ’= m x + b

           y ’= 0.2668 x + 0.00406

If we compare this equation of the line with the equation of free fall

          y ’= t2

          x = y

          m = 0.2668

          b = 0.00406

Since the value of b is very small, we can say that our assumption that the initial velocity is zero is correct.

Now we can compare the expressions of free fall and the equation of the line obtained

             y = ½ g t2

             y ’= m x

to be functionally equivalent

            x = 1 / m y '

now we can see the slope

          1 / m = ½ g

          g = 2 / m

          g = 2 / 0.2668

          g = 7.496 m / s²

b) the uncertainty of the acceleration is

          Δg = dg/dm   Δm

the uncertainty of m is given by the fit of the linear regression of the program Δm = 0.025

          dg / dm = 1 / 2m

           

           Δg = 1 /2m  Δm

           Δg = 1 / (2 0.2668) 0.025

           Δg = 0.047

The result of this exercise with the correct significant figures

          g = (0.13 + - 0.05) m / s²