# 4.17: Acceleration Due to Gravity

**At Grade**Created by: CK-12

**Practice**Acceleration Due to Gravity

In the absence of air resistance, all objects fall toward the earth with the same acceleration. Man, however, make maximum use of air resistance in the construction of parachutes for both entertainment and military use.

The image at left was taken during a 2008 Graduation demonstration jump by the U.S. Army Parachute Team. The 2008 team contained the first amputee member and the largest number of females in history.

### Acceleration Due to Gravity

One of the most common examples of uniformly accelerated motion is that an object allowed to fall vertically to the earth. 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*}*\begin{align*}g\end{align*} g* is 9.80 m/s

^{2}.

*All of the equations involving constant acceleration can be used for falling bodies but we insert \begin{align*}g\end{align*}*

^{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 \begin{align*}1.00 \ \text{s} = \left(\frac{1}{2}\right)(9.80 \ \text{m/s}^2)(1.00 \ \text{s})^2 = 4.90 \ \text{m}\end{align*}

Displacement after \begin{align*}2.00 \ \text{s} = \left(\frac{1}{2}\right)(9.80 \ \text{m/s}^2)(2.00 \ \text{s})^2 = 19.6 \ \text{m}\end{align*}

Displacement after \begin{align*}3.00 \ \text{s} = \left(\frac{1}{2}\right)(9.80 \ \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 -9.80 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)(9.80 \ \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 seconds. 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 seconds.

**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*} and we can also 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{ave}})(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 seen to hit 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)(9.80 \ \text{m/s}^2)(5.5 \ \text{s})^2 = 150 \ \text{m}\end{align*}

#### Summary

- 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 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 9.80 m/s^{2}*.*

#### Practice

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

http://www.youtube.com/watch?v=izXGpivLvgY

#### 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?

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### Image Attributions

- Solve problems of the motion of objects uniformly accelerated by gravity.

## Concept Nodes:

- acceleration due to gravity: The acceleration experienced by a body in free fall in a gravitational field.