In the absence of air resistance, all objects fall toward the Earth with the same acceleration. Parachutists, like the one from the U.S. Army Parachute Team shown above, make maximum use of air resistance in order to limit the acceleration of the fall.

### Acceleration Due to Gravity

One of the most common examples of uniformly accelerated motion is that an object allowed to drop will fall vertically to the Earth due to gravity. In treating falling objects as uniformly accelerated motion, we must ignore air resistance. Galileo’s original statement about the motion of falling objects is:

*At a given location on the Earth and in the absence of air resistance, all objects fall with the same uniform acceleration.*

We call this **acceleration due to gravity** on the Earth and we give it the symbol \begin{align*}g\end{align*}. The value of *\begin{align*}g\end{align*}* is 9.81 m/s^{2} in the downward direction. All of the equations involving constant acceleration can be used for falling bodies but we insert \begin{align*}g\end{align*} wherever “\begin{align*}a\end{align*}” appeared and the value of \begin{align*}g\end{align*} is always 9.81 m/s^{2}*.*

**Example: ** A rock is dropped from a tower 70.0 m high. How far will the rock have fallen after 1.00 s, 2.00 s, and 3.00 s? Assume the distance is positive downward.

**Solution: ** We are looking for displacement and we have time and acceleration. Therefore, we can use \begin{align*}d = \frac{1}{2} at^2\end{align*}.

Displacement after 1.00 s: \begin{align*}\left(\frac{1}{2}\right)(9.81 \ \text{m/s}^2)(1.00 \ \text{s})^2 = 4.91 \ \text{m}\end{align*}

Displacement after 2.00 s: \begin{align*}\left(\frac{1}{2}\right)(9.81 \ \text{m/s}^2)(2.00 \ \text{s})^2 = 19.6 \ \text{m}\end{align*}

Displacement after 3.00 s: \begin{align*}\left(\frac{1}{2}\right)(9.81 \ \text{m/s}^2)(3.00 \ \text{s})^2 = 44.1 \ \text{m}\end{align*}

**Example: ** (a) A person throws a ball upward into the air with an initial velocity of 15.0 m/s. How high will it go before it comes to rest? (b) How long will the ball be in the air before it returns to the person’s hand?

**Solution: ** In part (a), we know the initial velocity (15.0 m/s), the final velocity (0 m/s), and the acceleration (-9.81 m/s^{2}). We wish to solve for the displacement, so we can use \begin{align*}v{_f}^2 = v{_i}^2 + 2ad\end{align*} and solve for \begin{align*}d\end{align*}.

\begin{align*}d=\frac{v{_f}^2-v{_i}^2}{2a}=\frac{(0 \ \text{m/s})^2-(15.0 \ \text{m/s})^2}{(2)(-9.81 \ \text{m/s}^2)}=11.5 \ \text{m}\end{align*}

There are a number of methods by which we can solve part (b). Probably the easiest is to divide the distance traveled by the average velocity to get the time going up and then double this number since the motion is symmetrical—that is, time going up equals the time going down.

The average velocity is half of 15.0 m/s, or 7.5 m/s, and dividing this into the distance of 11.5 m yields 1.53 s. This is the time required for the ball to go up and the time for the ball to come down will also be 1.53 s, so the total time for the trip up and down is 3.06 s.

**Example: **A car accelerates with uniform acceleration from 11.1 m/s to 22.2 m/s in 5.0 s. (a) What was the acceleration and (b) how far did it travel during the acceleration?

**Solution: **

(a) \begin{align*}a=\frac{\Delta v}{\Delta t}=\frac{22.2 \ \text{m/s}-11.1 \ \text{m/s}}{5.0 \ \text{s}}=2.22 \ \text{m/s}^2\end{align*}

(b) We can find the distance traveled by \begin{align*}d = v_it + \frac{1}{2} at^2\end{align*}, or we could find the distance traveled by determining the average velocity and multiply it by the time.

\begin{align*}d &= v_it + \frac{1}{2} at^2 \\ &= (11.1 \ \text{m/s})(5.0 \ \text{s}) + \left(\frac{1}{2}\right)(2.22 \ \text{m/s}^2)(5.0 \ \text{s})^2 \\ &= 55.5 \ \text{m} + 27.8 \ \text{m} \\ &= 83 \ \text{m}\end{align*}

\begin{align*}d = (v_{\text{avg}})(t) = (16.6 \ \text{m/s})(5.0 \ \text{s}) = 83 \ \text{m}\end{align*}

**Example: ** A stone is dropped from the top of a cliff. It is hits the ground after 5.5 s. How high is the cliff?

**Solution: **

\begin{align*}d = v_it + \frac{1}{2} at^2 = (0 \ \text{m/s})(5.5 \ \text{s}) + \left(\frac{1}{2}\right)(9.81 \ \text{m/s}^2)(5.5 \ \text{s})^2 = 150 \ \text{m}\end{align*}

### Interactive Simulations

### Summary

- At any given location on the Earth and in the absence of air resistance, all objects fall with the same uniform acceleration.
- We call this acceleration the acceleration due to gravity on the Earth and we give it the symbol \begin{align*}g\end{align*}.
- The value of
*\begin{align*}g\end{align*}*is 9.81 m/s^{2}*.*

### Review

- A baseball is thrown vertically into the air with a speed of 24.7 m/s.
- How high does it go?
- How long does the round trip up and down require?

- A salmon jumps up a waterfall 2.4 m high. With what minimum speed did the salmon leave the water below to reach the top?
- A kangaroo jumps to a vertical height of 2.8 m. How long will it be in the air before returning to Earth?

### Explore More

This video offers a discussion and demonstration of the acceleration due to gravity.

- What is the gravitational acceleration given in the video? Why does it differ from that given in this text?
- Why does the ball travel further in later time intervals than in the earlier ones?