2.3 - Free Fall Motion

Free Fall and Gravitational Acceleration

Free fall occurs when gravity is the only force affecting an object’s motion; air resistance and friction are absent or neglected.
In air, resistance can cause objects to fall at different rates. In a vacuum, objects of different masses fall with the same acceleration: a feather and a ball bearing released together from the same height reach the floor together.
Gravitational acceleration, $g$ , is acceleration caused by gravitational force. Near Earth’s surface, its average value is $9.81 m s^{- 2}$ .
In a uniform gravitational field, a freely falling object’s downward velocity increases by $9.81 m s^{- 1}$ each second.
Galileo’s demonstrations established equal falling acceleration when air resistance is neglected, contrary to Aristotle’s view that falling rate depended on weight, shape and orientation.

Aim: determine Earth’s gravitational acceleration.
Apparatus: photogate system, electronic timer, electromagnetic release, steel ball and a container to catch the ball. The setup places two photogates below the release on a tripod stand with a height adjuster.

Position the second photogate 30.0 cm below the first. Align the ball so that it passes through both photogates into the container.
Release the steel ball using the electromagnetic release.
Record the elapsed times from release to the first and second photogates as $t_{1}$ and $t_{2}$ .
Repeat with photogate separations of 40.0, 50.0, 60.0 and 70.0 cm.
For each separation $h$ , calculate $g$ with $h$ in metres, then calculate the mean of the five values:

g = \frac{2 h}{t _{2}^{2} - t _{1}^{2}}

The recorded quantities are $h$ , $t_{1}$ , $t_{2}$ and calculated $g$ . Compare the mean with the equatorial value $9.78 m s^{- 2}$ .

Earth is not a perfect sphere. Its equatorial surface is farther from its centre than its polar surface.
At the equator, the distance to Earth’s centre is 6372 km and $g = 9.78 m s^{- 2}$ .
At the poles, the distance is 6350 km and $g = 9.83 m s^{- 2}$ .
The equatorial value is therefore smaller. The general value used for surface calculations is $9.81 m s^{- 2}$ .

For motion under gravity alone:

v = u + g t, s = u t + \frac{1}{2} g t^{2}, v^{2} = u^{2} + 2 g s

Taking upwards as positive, downward acceleration is $g = - 9.81 m s^{- 2}$ .
A ball thrown upwards has positive initial velocity; its velocity becomes zero at maximum height.
An object released from rest has $u = 0$ . Downward displacement and velocity are negative under this sign convention.

Amirah’s ball, thrown vertically upwards at $10 m s^{- 1}$ , neglecting air resistance:

0 = 10 - 9.81 t \Rightarrow t = 1.02 s

0 = 1 0^{2} + 2 (- 9.81) s \Rightarrow s = 5.10 m

Chan’s stone, released from a 10 m cliff, neglecting air resistance:

- 10 = \frac{1}{2} (- 9.81) t^{2} \Rightarrow t = 1.43 s

v^{2} = 2 (- 9.81) (- 10) \Rightarrow v = - 14.0 m s^{- 1}

Choose the positive time and the downward velocity; the stone reaches the ground moving downwards.