Free fall
Motion & Kinematics - OCR A-Level Physics
Key Definition
Free fall
Motion under the influence of gravity alone, with no other forces acting. The only acceleration is $g = 9.81 \text{ m s}^{-2}$ directed downward. On the Moon, $g = 1.6 \text{ m s}^{-2}$, so a hammer and feather released together hit the ground at the same time.
Motion under the influence of gravity alone, with no other forces acting. The only acceleration is $g = 9.81 \text{ m s}^{-2}$ directed downward. On the Moon, $g = 1.6 \text{ m s}^{-2}$, so a hammer and feather released together hit the ground at the same time.
Case 1: Dropped from rest
$u = 0$ and $a = +g$ (taking downward as positive). Time to fall a height $h$ is $t = \sqrt{\frac{2h}{g}}$. Speed on impact is $v = \sqrt{2gh}$.
Case 2: Projected upward
Take upward as positive. $u > 0$, $a = -g$. Time to highest point: $t = u/g$. The flight is symmetric: time up equals time down. The object returns to launch height with speed $u$ in the downward direction.
- In free fallMotion under gravity alone, with no other forces acting. All objects in free fall near Earth's surface have the same acceleration, g = 9.81 m s⁻²., all objects accelerate at the same rate regardless of mass (in the absence of air resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).. Measured in ohms (Ω).)
- g = 9.81 $m s^{-2}$ near the Earth's surface, directed vertically downwards
- When using SUVAT for vertical motion, choose a sign convention (e.g. $upwards = positive)$ and be consistent throughout
- An object thrown upwards decelerates at g; at the highest point v = 0; it then accelerates downwards at g
- The time to rise to the highest point equals the time to fall back to the launch height (ignoring air resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to current. Measured in ohms (Ω).)
Common Mistake
MEDIUM
Students often: Don't forget assign a negative sign to g when taking upward as positive, leading to wrong answers for objects thrown upwards.
Instead: If upward is positive, then a = -9.81 $m s^{-2}$ for all free-fall problems. If downward is positive, a = +9.81 $m s^{-2}$. State your convention clearly.
Instead: If upward is positive, then a = -9.81 $m s^{-2}$ for all free-fall problems. If downward is positive, a = +9.81 $m s^{-2}$. State your convention clearly.
Worked Example [3 marks]
A stone is dropped from rest from a bridge 45 m above a river. Calculate the time taken to reach the water and the speed on impact. Use g = 9.81 $m s^{-2}$.
Show Solution
1
Known
$s = 45 \text{ m}, u = 0, a = 9.81 \text{ m s}^{-2}. Use s = ut + \frac{1}{2}at^{2}: 45 = 0 + \frac{1}{2}(9.81)t^{2}$
[1]2
$t^{2} = \frac{2 \times 45}{9.81} = 9.174 \Rightarrow t = 3.03 \text{ s} \approx 3.0 \text{ s}$
[1]3
$Use v = u + at: v = 0 + 9.81 \times 3.03 = 29.7 \text{ m s}^{-1} \approx 30 \text{ m s}^{-1}$
[1]Answer
Time = 3.0 s; speed on impact = 30 $m s^{-1}$
Diagram pending
Free-fall apparatus. Electromagnet at top holds a steel ball; releasing the switch starts a timer and drops the ball. A hinged trapdoor at the bottom is struck by the ball and stops the timer. Height $h$ marked between the ball and the trapdoor. Second panel: graph of $h$ on the y-axis against $t^2$ on the x-axis, straight line through origin with gradient $g/2$.
Will be replaced with a GeoGebra SVG in stream 2.
Experiment: determining $g$ by free fall
Apparatus: electromagnet and trapdoor timerAn electromagnet releases a steel ball; a trapdoor at the bottom stops the timer on impact., steel ball, metre rule, timer reading to $0.01 \text{ s}$.
Method: measure several heights $h$, time each fall $t$, repeat for each height, take the mean $t$. Calculate $t^2$. Plot $h$ on the $y$-axis against $t^2$ on the $x$-axis.
Analysis: From $h = \tfrac{1}{2} g t^2$, comparing with $y = mx + c$, the gradient equals $g/2$. Multiply the gradient by $2$ to get $g$.
Method: measure several heights $h$, time each fall $t$, repeat for each height, take the mean $t$. Calculate $t^2$. Plot $h$ on the $y$-axis against $t^2$ on the $x$-axis.
Analysis: From $h = \tfrac{1}{2} g t^2$, comparing with $y = mx + c$, the gradient equals $g/2$. Multiply the gradient by $2$ to get $g$.
- Worked example (Oxford data). Heights $1.18, 1.00, 0.84, 0.74, 0.62, 0.54 \text{ m}$ give times $0.482, 0.447, 0.409, 0.386, 0.353, 0.325 \text{ s}$. Plotting $h$ vs $t^2$ gives a gradient near $5.1 \text{ m s}^{-2}$, so $g \approx 10.2 \text{ m s}^{-2}$.
- Why plot $h$ vs $t^2$ rather than calculate from one reading? A best-fit line averages the random error across all points and the gradient is more reliable than any single $g = 2h/t^2$ calculation.
- Alternative apparatus: light gatesDevices that detect when an object passes through, used to measure speed by timing across a known distance. measure $u$ and $v$ over a known separation $s$; then $g = (v^2 - u^2)/(2s)$.
- Sources of error: human reaction time on hand-held stopwatches (eliminated by electronic timer), air resistance (small for a dense steel ball over short drops), measurement of $h$ (use a metre rule with clear endpoints).
Examiner Tips and Tricks
- In practical questions about measuring g, always describe the graph you would plot.
- Plotting s against \(t^{2}\) gives a straight line through the origin with gradient g/2.
- This is far more reliable than a single measurement.