Earth’s gravitational acceleration
Near the surface of the Earth, the acceleration due to gravity is approximately constant, with a magnitude of about
\[\boxed{g = 9.8\ \mathrm{m/s^2}.}\]All objects near Earth’s surface experience this same acceleration.
This is the acceleration of objects that are subject only to the gravitational force, and not other forces like air resistance. We call motion under gravity alone free-fall motion. The direction of the gravitational acceleration is downward, toward the center of the Earth.
Free-fall motion
Be careful with these common points of confusion about free-fall motion:
Key result:
- $g$ is the magnitude of the gravitational acceleration near Earth’s surface. Because it is the magnitude of a vector, $g$ is always positive. Any time you see the symbol $g$, it refers to the positive number $9.8\ \mathrm{m/s^2}$.
- Being in free fall just means that the only force acting on the object is gravity. It does not necessarily mean that the object is falling downward. For example, a ball thrown straight up in the air is in free-fall motion, even though it is moving upward at first.
We can only apply the free-fall model to objects that are moving freely under the influence of gravity. If an object is in contact with other objects, like the ground or a person’s hand, then it is not in free fall.
Free-fall motion is constant acceleration motion
Since the free-fall acceleration is constant, free-fall motion is a special case of constant acceleration motion. We can apply the same constant acceleration equations we just learned:
\[\begin{aligned} y(t) &= y_0 + v_{0y} t + \frac{1}{2} a_y t^2\\ v_y(t) &= v_{0y} + a_y t\\ v_{f,y}^2 &= v_{i,y}^2 + 2 a_y \Delta y. \end{aligned}\]When modeling free-fall motion, I will use the convention that the $y$-axis is vertical, with the positive $y$ direction pointing up. Since the gravitational acceleration points down, the $y$-component of the acceleration is negative:
\[\boxed{a_y = -g}\]Plugging this free-fall acceleration $a_y = -g$ into the constant acceleration equations gives the following equations for free-fall motion:
\[\begin{aligned} y(t) &= y_0 + v_{0y} t + \frac{1}{2} (-g) t^2\\ &= y_0 + v_{0y} t - \frac{1}{2} g t^2\\ v_y(t) &= v_{0y} + (-g) t\\ &= v_{0y} - g t\\ v_{f,y}^2 &= v_{i,y}^2 + 2 (-g) \Delta y\\ &= v_{i,y}^2 - 2 g \Delta y \end{aligned}\]Check your understanding: A ball is held in your hand, thrown upward, released, rises to its highest point, then falls back down and hits the ground. (a) During which parts of this motion is the ball in free fall? (b) For the free-fall parts, what are the signs of $v_y$ and $a_y$ if the positive $y$-direction points upward?
Answer
(a) An object is in free fall whenever gravity is the only force acting on it. While the ball is still in your hand, your hand is also pushing on it, so it is not in free fall during the throwing phase. The instant the ball leaves your hand, gravity is the only force acting, and it stays that way while the ball rises, at the highest point, and while it falls. So the ball is in free fall from the moment it leaves your hand until just before it hits the ground.
(b) With $+y$ pointing up, the free-fall acceleration is
\[a_y = -g = -9.8\ \mathrm{m/s^2}\]at every instant of the free-fall motion. The acceleration is the same on the way up, at the top, and on the way down. The velocity, however, changes sign:
- On the way up: the ball moves upward, so $v_y > 0$.
- At the highest point: the ball is momentarily at rest in the vertical direction, so $v_y = 0$ — but $a_y$ is still $-g$ (which is why the ball does not stay there).
- On the way down: the ball moves downward, so $v_y < 0$.
A common mistake is to think that $a_y = 0$ at the top because $v_y = 0$ there. But zero velocity does not mean zero acceleration: the velocity is still changing (from positive to negative) at that instant, so the acceleration is nonzero.
Example: Free-fall motion of a tossed ball
A ball is tossed straight up from the edge of a $10.0\ \mathrm{m}$ high building at $9.00\ \mathrm{m/s}$.
- At the top of its motion, what is the ball’s height above the building?
- When the ball returns to its initial height, how fast is it moving?
- When the ball hits the ground, how long has it been in the air?
- Just before it hits the ground, how fast is the ball moving?
Solution
Take the $+y$ axis pointing upward, with the origin $y = 0$ at the ground. The ball starts at height $y_0 = 10.0\ \mathrm{m}$ with initial velocity $v_0 = +9.00\ \mathrm{m/s}$ (positive because it is tossed upward). Once released, the ball is in free fall, so we can use the free-fall kinematic equations:
\[\begin{aligned} y(t) &= y_0 + v_0\,t - \tfrac{1}{2} g t^2,\\ v_y(t) &= v_0 - g t,\\ v_{y,f}^2 &= v_0^2 - 2 g\,\Delta y, \end{aligned}\]with $g = 9.8\ \mathrm{m/s^2}$.
(a) At the maximum height, the ball’s velocity is instantaneously zero ($v_y = 0$). Using the $v^2$ equation with $v_y = 0$,
\[\begin{aligned} 0 &= v_0^2 - 2 g\,h_{\max}\\ \rightarrow h_{\max} &= \frac{v_0^2}{2g} = \frac{(9.00\ \mathrm{m/s})^2}{2\,(9.8\ \mathrm{m/s^2})} = \boxed{4.13\ \mathrm{m}.} \end{aligned}\]This is the height above the building (above the launch point).
(b) When the ball returns to its initial height, its displacement from the start is $\Delta y = 0$. Using the $v^2$ equation,
\[v_{y,f}^2 = v_0^2 - 2 g (0) = v_0^2.\]Taking the square root, the speed is simply the initial speed, $\boxed{9.00\ \mathrm{m/s}.}$ This is a general fact about free fall: the speed at a given height is the same on the way up as on the way down.
(c) The ball reaches the ground when $y = 0$, so
\[y_0 + v_0\,t - \tfrac{1}{2} g t^2 = 0.\]This is a quadratic equation of the form $A t^2 + B t + C = 0$ with
\[A = -\tfrac{1}{2} g, \qquad B = v_0, \qquad C = y_0.\]Using the quadratic formula and simplifying,
\[t = \frac{-B \pm \sqrt{B^2 - 4 A C}}{2 A} = \frac{v_0 \pm \sqrt{v_0^2 + 2 g y_0}}{g}.\]Plugging in the numbers,
\[t = \frac{9.00\ \mathrm{m/s} \pm \sqrt{(9.00\ \mathrm{m/s})^2 + 2\,(9.8\ \mathrm{m/s^2})(10.0\ \mathrm{m})}}{9.8\ \mathrm{m/s^2}} = \begin{cases} \phantom{-}2.62\ \mathrm{s} & (+)\\ -0.780\ \mathrm{s} & (-) \end{cases}\]We want the positive time after the toss, so the ball is in the air for $\boxed{2.62\ \mathrm{s}.}$
(d) The quickest route is the $v^2$ equation with $\Delta y = -y_0$ (the ball ends up a distance $y_0$ below its start):
\[\begin{aligned} v_{y,f}^2 &= v_0^2 - 2 g (-y_0) = v_0^2 + 2 g y_0\\ \rightarrow \lvert v_{y,f} \rvert &= \sqrt{v_0^2 + 2 g y_0} = \sqrt{(9.00\ \mathrm{m/s})^2 + 2\,(9.8\ \mathrm{m/s^2})(10.0\ \mathrm{m})} = \boxed{16.6\ \mathrm{m/s}.} \end{aligned}\]As a check, plugging $t = 2.62\ \mathrm{s}$ into the velocity equation gives $v_y = 9.00\ \mathrm{m/s} - (9.8\ \mathrm{m/s^2})(2.62\ \mathrm{s}) = -16.6\ \mathrm{m/s}$, whose magnitude is the same speed.