AshPhys

Acceleration caused by gravity

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.2Chapter 2: Accelerated motionLesson 2.11Sections 2.11

Goal: Use g = 9.81 m s⁻² in the suvat equations for objects falling, or thrown up, without air resistance.

01The big picture

Near the Earth’s surface, with no air resistance, every object falls with the same acceleration g=9.81g = 9.81 m s⁻², downwards, whatever its mass. This is free fall.

An object thrown upwards has the same downward acceleration all the way: it slows, stops for an instant at the top (where v=0v = 0 but aa is still gg) and speeds up on the way down. Take up as positive and use a=−9.81a = -9.81 m s⁻², or down as positive and a=+9.81a = +9.81.

Acceleration of free fall gg [paraphrase]
The acceleration of a falling object with no air resistance: 9.81 m s⁻² near the Earth.
Fig. 1 · v–t for a ball thrown up
v–t for a ball thrown upVelocity against time, up positive. A straight line falls from +12 m/s at time 0 through zero at 1.22 s (the top of the flight) to −12 m/s at 2.45 s.v / m s⁻¹t / stop: v = 0, a = −g12−120going upcoming down
One straight line with gradient −g-g all the way, through v=0v = 0 at the top.

02Key equations

Dropped from rest: time
s=12gt2s = \tfrac{1}{2}gt^{2}
Dropped from rest: speed
v2=2gsv^{2} = 2gs
Symbols and units
  • ggacceleration of free fall, 9.81m s−2\mathrm{m\,s^{-2}}
  • ssdistance fallenm\mathrm{m}

03Sample problems with solutions

Sample 1 · A dropped stone
A stone is dropped from rest from a bridge 20 m above the water. Ignoring air resistance, calculate the time it takes to fall and its speed when it hits the water.

  1. Down positive, u=0u = 0, a=9.81a = 9.81:

    t=2sg=2×209.81=2.02 st = \sqrt{\frac{2s}{g}} = \sqrt{\frac{2 \times 20}{9.81}} = 2.02\ \mathrm{s}

  2. Speed:

    v=2gs=2×9.81×20=19.8 m s−1v = \sqrt{2gs} = \sqrt{2 \times 9.81 \times 20} = 19.8\ \mathrm{m\,s^{-1}}
Answer: 2.0 s; 20 m s⁻¹ (19.8 m s⁻¹)
Unit check: m s−2×m=m s−1 ✓\sqrt{\mathrm{m\,s^{-2}} \times \mathrm{m}} = \mathrm{m\,s^{-1}}\ \checkmark

04Practice problems

  1. P1
    A ball is thrown up at 12 m s⁻¹. How long does it take to reach the top?
    Final answer1.2 s (1.22 s)
    Answer: 1.2 s (1.22 s)
  2. P2
    How high does that ball rise?
    Final answer7.3 m
    Answer: 7.3 m
  3. P3
    A ball is dropped from rest. Find its speed and the distance fallen after 3.0 s.
    Final answer29 m s⁻¹ (29.4); 44 m (44.1)
    Answer: 29 m s⁻¹ (29.4); 44 m (44.1)
  4. P4
    A hammer and a feather are dropped together on the Moon (no air). Which lands first?
    Final answerNeither: they land together
    Answer: Neither: they land together

05Other useful information

Common mistakes

  • ✗ Wrong: At the top of its flight the acceleration is zero.

    ✓ Right: Only vv is zero; aa is still 9.81 m s⁻² downwards.

Exam tips

  • Write your sign convention (“up +”) at the top of the working.

What the syllabus asks

  • 9702 topic 2: free fall without air resistance. Coursebook 2.11.

Explore: Practice questions for this lesson