Equation of motion for free fall

A body in free fall moves under the force of gravity alone. Its velocity does not stay the same. Near the surface of the earth it increases by about 9.8 m/s in every second of the fall. An increase of the same amount every second is a uniform acceleration, and uniform acceleration is exactly the kind of motion the equations of motion describe. So the equations that work for a bus speeding up on a straight road also work for a stone falling from a roof.

Because we already know the acceleration of a falling body, we can work out three things about its fall that are hard to measure directly:

• the velocity with which it reaches the ground,
• the time it takes to reach the ground,
• the height from which it was dropped.

For a body moving in a straight line with uniform acceleration \( a \), three equations connect its initial velocity \( u \), final velocity \( v \), time \( t \) and distance \( s \). In free fall the acceleration is produced by the force of gravity, so its value is the acceleration due to gravity, \( g \). We simply write \( g \) in place of \( a \). The distance a falling body covers is a height, so we also write \( h \) in place of \( s \).

Each letter stands for one quantity, and each quantity has its own unit:

• \( u \) is the initial velocity, measured in m/s.
• \( v \) is the final velocity, measured in m/s.
• \( g \) is the acceleration due to gravity, \( 9.8 \text{ m/s}^2 \) near the surface of the earth.
• \( t \) is the time taken, measured in s.
• \( h \) is the height, measured in m.

Gravity always pulls towards the centre of the earth, which means downward. Whether that pull speeds a body up or slows it down depends on which way the body is moving.

When a body is dropped, it moves downward, in the same direction as the pull of gravity. Its velocity increases, so the acceleration is positive and we use \( g = +9.8 \text{ m/s}^2 \).

When a body is thrown vertically upward, it moves against the pull of gravity. Gravity now acts in the opposite direction to the motion and slows the body down. The acceleration is therefore negative, and we use \( g = -9.8 \text{ m/s}^2 \). A ball going up loses 9.8 m/s of velocity in every second. Its velocity keeps dropping until it becomes zero, and at that instant the ball is at the highest point of its path. It then starts to fall back, and on the way down its velocity grows again.

A stone is dropped from a bridge into the river below. The splash is heard 2 seconds after the stone is let go. Find the height of the bridge above the water. Take \( g = 9.8 \text{ m/s}^2 \).

Over a short height like this, the sound reaches our ears almost at once, so the 2 seconds is taken as the time of the fall.

What we know:

• Initial velocity \( u = 0 \) m/s, because the stone is dropped from the hand, not thrown.
• Time of fall \( t = 2 \) s.
• Acceleration \( g = +9.8 \text{ m/s}^2 \), positive because the stone moves downward.
• Height \( h = ? \)

Step 1: The equation that connects height with time is:

\[ h = ut + \frac{1}{2}gt^2 \]

Step 2: Putting in the values. Since \( u = 0 \), the term \( ut \) is zero:

\[ h = 0 \times 2 + \frac{1}{2} \times 9.8 \times 2^2 \]

Step 3: Squaring the time, \( 2^2 = 4 \), gives:

\[ h = \frac{1}{2} \times 9.8 \times 4 \]

Step 4: Multiplying out:

\[ h = 19.6 \text{ m} \]

So the bridge is 19.6 m above the surface of the water.

A stone dropped from the roof of a building takes 3 s to reach the ground. Find the velocity with which it hits the ground and the height of the roof. Take \( g = 9.8 \text{ m/s}^2 \).

What we know:

• \( u = 0 \) m/s, because the stone is dropped.
• \( t = 3 \) s.
• \( g = +9.8 \text{ m/s}^2 \), because the stone moves downward.
• \( v = ? \) and \( h = ? \)

Step 1: The equation that gives velocity from time is:

\[ v = u + gt \]

Step 2: Putting in \( u = 0 \), \( g = 9.8 \) and \( t = 3 \):

\[ v = 0 + 9.8 \times 3 \]

Step 3: Multiplying:

\[ v = 29.4 \text{ m/s} \]

Step 4: For the height, the equation that uses time is:

\[ h = ut + \frac{1}{2}gt^2 \]

Step 5: Putting in the values, with \( 3^2 = 9 \):

\[ h = 0 + \frac{1}{2} \times 9.8 \times 9 \]

Step 6: Multiplying:

\[ h = 44.1 \text{ m} \]

The stone hits the ground at 29.4 m/s, and the roof is 44.1 m high.

The third equation gives a check. In \( v^2 = u^2 + 2gh \), the right side is \( 2 \times 9.8 \times 44.1 = 864.36 \), and \( 29.4^2 \) is also 864.36. The answers agree because all three equations describe the same fall.

Look at how the velocity grows: 9.8 m/s after 1 s, 19.6 m/s after 2 s and 29.4 m/s after 3 s. The velocity increases uniformly. The distance covered in each second does not: the stone falls 4.9 m in the first second, 14.7 m in the second and 24.5 m in the third, because it is moving faster in every later second.

A cricket ball is thrown vertically upward and reaches a height of 30 m. Find the velocity with which it was thrown and the time it takes to reach the maximum height.

What we know:

• Maximum height \( h = 30 \) m.
• Final velocity \( v = 0 \) m/s, because the ball stops for an instant at the maximum height.
• Acceleration \( g = -9.8 \text{ m/s}^2 \), negative because the ball moves up against gravity.
• Initial velocity \( u = ? \) and time \( t = ? \)

First, the velocity of the throw. The time is not given, so we choose the equation that has no t in it.

Step 1: The equation without time is:

\[ v^2 = u^2 + 2gh \]

Step 2: Putting in \( v = 0 \), \( g = -9.8 \) and \( h = 30 \):

\[ 0 = u^2 + 2 \times (-9.8) \times 30 \]

Step 3: Multiplying the last term. Since \( 2 \times 9.8 \times 30 = 588 \), this becomes:

\[ 0 = u^2 - 588 \]

Step 4: Moving 588 to the other side:

\[ u^2 = 588 \]

Step 5: Taking the square root of both sides:

\[ u = \sqrt{588} \approx 24.25 \text{ m/s} \]

Next, the time to reach the top. Now that u is known, we can use the equation linking velocity and time.

Step 6: The equation linking velocity and time is:

\[ v = u + gt \]

Step 7: Putting in \( v = 0 \), \( u = 24.25 \) and \( g = -9.8 \):

\[ 0 = 24.25 - 9.8 \times t \]

Step 8: Moving the time term to the left side:

\[ 9.8 \times t = 24.25 \]

Step 9: Dividing both sides by 9.8:

\[ t = \frac{24.25}{9.8} \approx 2.47 \text{ s} \]

The ball was thrown upward at about 24.25 m/s, and it takes about 2.47 s to reach its maximum height of 30 m.

Before calculating anything, write down the five quantities u, v, g, t and h. Fill in the ones the question gives, mark the one it asks for, and then choose the equation that contains the unknown together with quantities you already know. Each of the three equations leaves out one quantity, and that is the quickest way to pick between them.

Some problems need two equations, one after the other. In the cricket ball example, the velocity found from the first equation was then used in the second to find the time.

These equations let you find the height of a house or a school building without climbing up with a tape. A stone is dropped from the roof, and a tin placed on the ground below makes a sound at the exact moment the stone lands. The time of the fall is measured, and because \( u = 0 \) the height comes straight from \( h = \frac{1}{2}gt^2 \). For example, a fall that takes 1.5 s gives \( h = \frac{1}{2} \times 9.8 \times 1.5^2 \), which is 11.025 m.

A timed fall is never perfectly exact. Starting and stopping a watch by hand takes a fraction of a second, and the air pushes back a little on the stone. So the fall is timed several times from the same height and the average time is used, which reduces the probable error. Measuring the actual height with a measuring tape then shows how good the calculation was. The difference between the calculated height and the measured height is the error in the calculated height.