What is free fall? When an object falls toward Earth with only gravity acting on it (no push, no air resistance), we say it is in free fall. During free fall the object's velocity keeps increasing at a steady rate — it accelerates. That acceleration is exactly the acceleration due to gravity, $g \approx 9.8\ \text{m/s}^2$. A famous demonstration: in a vacuum tube, a feather and a hammer dropped together hit the bottom at the same instant, because both fall with the same $g$ regardless of mass.
Equations of motion under gravity: Free fall is just uniformly accelerated motion with $a = g$. So the three equations of motion you already know become, taking downward as positive:
- $v = u + gt$
- $h = ut + \dfrac{1}{2} g t^2$
- $v^2 = u^2 + 2 g h$
Here $u$ is the initial velocity, $v$ the final velocity, $h$ the height (or distance) fallen, and $t$ the time. If an object is simply dropped, $u = 0$, which simplifies these nicely (for example $h = \dfrac{1}{2} g t^2$). If an object is thrown upward, take $g$ as a deceleration (negative), and at the highest point $v = 0$.
Mass: The mass of a body is the amount of matter it contains. It is a measure of the body's inertia — how hard it is to change its motion. Mass is measured in kilograms (kg), is a scalar, and is the same everywhere in the universe. A $5\ \text{kg}$ bag of rice is $5\ \text{kg}$ on Earth, on the Moon, or in deep space.
Weight: The weight of a body is the force with which a planet attracts it. Since weight is a force caused by gravity, $W = mg$. Weight is measured in newtons (N), is a vector (it points downward), and changes with location because $g$ changes. A $5\ \text{kg}$ bag weighs about $49\ \text{N}$ on Earth but far less on the Moon.
Mass vs weight — the key contrast:
- Mass is matter (kg, scalar, constant); weight is a force (N, vector, varies with $g$).
- Mass can never be zero for a real object; weight can be zero where $g = 0$ (deep space) and feels zero in free fall (the ‘weightlessness’ astronauts feel).
Weight on the Moon: The Moon has much less mass and a smaller radius than Earth, giving it a surface gravity of about $g_{moon} = 1.63\ \text{m/s}^2$, which is close to $\dfrac{1}{6}$ of Earth's $g$. So an object's weight on the Moon is roughly one-sixth of its weight on Earth, even though its mass is unchanged. This is why astronauts can leap so high in lunar footage — same muscles, same mass, but only one-sixth the weight to lift.
A stone is dropped from rest from a tower and falls for $3\ \text{s}$. Find its velocity just before hitting the ground and the height of the tower. (Take $g = 9.8\ \text{m/s}^2$.)
Solution- Dropped from rest, so $u = 0$, $t = 3\ \text{s}$, $g = 9.8\ \text{m/s}^2$.
- Velocity: $v = u + gt = 0 + 9.8 \times 3 = 29.4\ \text{m/s}$.
- Height: $h = ut + \dfrac{1}{2} g t^2 = 0 + \dfrac{1}{2}(9.8)(3)^2$.
- $h = \dfrac{1}{2}(9.8)(9) = 44.1\ \text{m}$.
Answer: Final velocity $= 29.4\ \text{m/s}$; tower height $= 44.1\ \text{m}$.
A ball is thrown vertically upward with a speed of $20\ \text{m/s}$. How high does it rise, and how long does it take to reach the top? (Take $g = 10\ \text{m/s}^2$.)
Solution- Going up, gravity decelerates the ball, so use $g = -10\ \text{m/s}^2$ with $u = 20\ \text{m/s}$.
- At the top $v = 0$. Use $v^2 = u^2 + 2gh$: $0 = (20)^2 + 2(-10)h$.
- $0 = 400 - 20h \Rightarrow h = \dfrac{400}{20} = 20\ \text{m}$.
- Time to top: $v = u + gt \Rightarrow 0 = 20 + (-10)t \Rightarrow t = 2\ \text{s}$.
Answer: Maximum height $= 20\ \text{m}$; time to reach the top $= 2\ \text{s}$.
Distinguish between mass and weight, giving two differences.
Solution- Mass is the amount of matter in a body; weight is the gravitational force on it ($W = mg$).
- Mass is measured in kilograms and is a scalar; weight is measured in newtons and is a vector.
- Mass is the same everywhere; weight changes with $g$ (location).
Answer: Mass = matter (kg, scalar, constant); Weight = force of gravity (N, vector, varies with location).
An object weighs $60\ \text{N}$ on Earth. Find (a) its mass and (b) its weight on the Moon, where $g_{moon} = \dfrac{1}{6} g_{earth}$. (Take $g = 10\ \text{m/s}^2$ on Earth.)
Solution- On Earth $W = mg \Rightarrow 60 = m \times 10 \Rightarrow m = 6\ \text{kg}$.
- Mass is the same on the Moon, so $m = 6\ \text{kg}$ there too.
- Moon's $g = \dfrac{1}{6} \times 10 = 1.67\ \text{m/s}^2$ (approx).
- Weight on Moon $= mg_{moon} = 6 \times 1.67 = 10\ \text{N}$ (i.e. $\dfrac{60}{6}$).
Answer: Mass $= 6\ \text{kg}$ (same everywhere); weight on the Moon $= 10\ \text{N}$, one-sixth of its Earth weight.
A stone is dropped from a height of $80\ \text{m}$. Find the time it takes to reach the ground. (Take $g = 10\ \text{m/s}^2$.)
Solution- Dropped, so $u = 0$, $h = 80\ \text{m}$, $g = 10\ \text{m/s}^2$.
- Use $h = ut + \dfrac{1}{2} g t^2 = \dfrac{1}{2} g t^2$.
- $80 = \dfrac{1}{2}(10) t^2 = 5 t^2$.
- $t^2 = 16 \Rightarrow t = 4\ \text{s}$.
Answer: The stone takes $4\ \text{s}$ to reach the ground.
Why do a coin and a feather, dropped together inside an evacuated (air-free) jar, reach the bottom at the same time?
Solution- In free fall the only force is gravity, giving acceleration $g = \dfrac{GM}{R^2}$.
- This $g$ is the same for every object and does not depend on mass.
- In ordinary air, the feather is slowed by air resistance, so it lags behind the coin.
- Remove the air (evacuated jar) and there is no air resistance, so both accelerate equally at $g$.
Answer: With no air resistance both fall with the same acceleration $g$, so they land together regardless of mass.