Free fall: bodies dropped and thrown upwards: WASSCE Physics

Drop a stone and a sheet of paper together: the stone wins. Crumple the paper into a tight ball and they land together. Air, not mass, made the difference. Without air, every object falls with exactly the same acceleration.

Learning objectives

By the end of this lesson you should be able to:

define acceleration due to gravity and free fall
apply the equations of motion to bodies falling or thrown vertically
find the maximum height and time of flight of a body thrown upwards
interpret the velocity–time graph of vertical motion

Free fall and the acceleration due to gravity

Definition

A body is in free fall when the only force acting on it is its weight (air resistance is negligible). It then falls with a constant acceleration called the acceleration due to gravity, $g$ .

Near the Earth's surface $g \approx 9.8$ m s⁻². WASSCE questions usually say take $g = 10$ m s⁻²; use whatever value the question gives.

Key facts about free fall:

$g$ is the same for all bodies at the same place, whatever their mass.
$g$ always acts vertically downwards, whether the body is moving up or down.
$g$ varies slightly over the Earth: larger at the poles than at the equator, smaller at the top of a mountain.

Common mistake

"Heavy things fall faster" is false in a vacuum. A heavier body has a bigger weight, but it also has more mass to accelerate; the two effects cancel ( $a = \dfrac{mg}{m} = g$ ). Air resistance is what slows light, broad objects such as a feather or a leaf.

The equations of motion for vertical motion

Replace $a$ with $\pm g$ and $s$ with height $h$ . Choose a direction as positive and stick to it.

Taking downwards as positive (convenient for dropped bodies):

v = u + gt \qquad h = ut + \tfrac{1}{2}gt^2 \qquad v^2 = u^2 + 2gh

\begin{gathered}v = u + gt \\[6pt] h = ut + \tfrac{1}{2}gt^2 \\[6pt] v^2 = u^2 + 2gh\end{gathered}

Taking upwards as positive (convenient for bodies thrown up): use $a = -g$ :

v = u - gt \qquad h = ut - \tfrac{1}{2}gt^2 \qquad v^2 = u^2 - 2gh

\begin{gathered}v = u - gt \\[6pt] h = ut - \tfrac{1}{2}gt^2 \\[6pt] v^2 = u^2 - 2gh\end{gathered}

A body dropped from rest

Worked example

A coconut falls from rest from a tree 20 m high. Find (a) the time to reach the ground, (b) its speed on impact. Take $g = 10$ m s⁻².

Downwards positive, $u = 0$ , $h = 20$ m.

(a) $h = \tfrac{1}{2}gt^2 \Rightarrow 20 = 5t^2 \Rightarrow t = 2$ s.

(b) $v = gt = 10 \times 2 = 20$ m s⁻¹ (or $v^2 = 2gh = 400$ ).

A body thrown vertically upwards

Velocity-time graph for a ball thrown straight up: a straight line falling from +u through zero at the highest point to -u when it returns

Figure 1. Velocity–time graph for a body thrown upwards (up positive). WASSCE Tutor diagram.

At the highest point the velocity is momentarily zero (but the acceleration is still $g$ downwards). Setting $v = 0$ :

Quantity	Result	How it is found
Time to reach the top	$t = \dfrac{u}{g}$	$0 = u - gt$
Maximum height	$H = \dfrac{u^2}{2g}$	$0 = u^2 - 2gH$
Time of flight (back to the thrower)	$T = \dfrac{2u}{g}$	twice the time up
Speed on return to the thrower	$u$ (same as launch speed)	symmetry, no air resistance

Worked example

A ball is thrown vertically upwards at 30 m s⁻¹. Find (a) the maximum height, (b) the time to reach it, (c) the total time in the air, (d) its velocity after 4 s. Take $g = 10$ m s⁻².

Up positive.

(a) $H = \dfrac{u^2}{2g} = \dfrac{900}{20} = 45$ m.

(b) $t = \dfrac{u}{g} = \dfrac{30}{10} = 3$ s.

(d) $v = u - gt = 30 - 10 \times 4 = -10$ m s⁻¹. The minus sign means it is now moving downwards at 10 m s⁻¹.

Common mistake

At the top of its flight, $v = 0$ but $a \neq 0$ . If the acceleration were zero there, the ball would stay in the air. Objective questions test this often: "At the highest point, the acceleration is…" — answer: $g$ , downwards.

Energy view (a useful check)

Ignoring air resistance, kinetic energy at launch becomes potential energy at the top:

\tfrac{1}{2}mu^2 = mgH \Rightarrow H = \dfrac{u^2}{2g}

\begin{gathered}\tfrac{1}{2}mu^2 = mgH \\[6pt] \Rightarrow H = \dfrac{u^2}{2g}\end{gathered}

the same result, with the mass cancelling.

Terminal velocity

In air, a falling body feels air resistance that grows with its speed. Eventually air resistance equals the weight; the resultant force is then zero and the body falls at a constant terminal velocity. A skydiver reaches about 55 m s⁻¹; opening a parachute raises the air resistance and lowers the terminal velocity to about 5 m s⁻¹.

In Sierra Leone

A raindrop falling from a cloud 1000 m up would hit the ground at about 140 m s⁻¹ if there were no air, enough to injure. Air resistance gives raindrops a terminal velocity of only a few metres per second, which is why the rainy season in Freetown is wet but not dangerous.

Key points

Free fall: only weight acts; acceleration $g$ (take 10 m s⁻² unless told otherwise), the same for all masses.
Use the equations of motion with $a = \pm g$ ; choose a positive direction and keep to it.
Thrown up: $t_{top} = u/g$ , $H = u^2/2g$ , time of flight $= 2u/g$ .
At the top: $v = 0$ , $a = g$ downwards.
With air resistance, a falling body reaches a constant terminal velocity.

WASSCE examination focus

WAEC tests this

Objective items test "the acceleration at the highest point", time of flight and maximum height. Theory items combine vertical motion with energy, or ask for the velocity–time graph of a ball thrown up and caught (a single straight line crossing the time axis, not a "V" shape). State your sign convention at the start.

Quick check

A stone is dropped from a bridge and hits the water 3 s later. How high is the bridge? ( $g = 10$ m s⁻²)
A ball is thrown up at 20 m s⁻¹. What is its maximum height?
What is the velocity of the ball in question 2 after 3 s?

Answers: 1. $h = \tfrac{1}{2}(10)(9) = 45$ m. 2. $H = \dfrac{400}{20} = 20$ m. 3. $v = 20 - 30 = -10$ m s⁻¹: 10 m s⁻¹ downwards.

Free fall: bodies dropped and thrown upwards