prev   2-1 - 2-3   2-4 - 2-6   2-7        

2-7 Freely Falling Objects

One application of constant acceleration is found when objects fall near Earth's surface. It is an experimental fact that when air resistance is negligible, object's near Earth's surface fall with a constant acceleration g. The symbol "g" represents the magnitude of this acceleration, which has an average value of

g = 9.81 m/s².

Objects undergoing this type of motion, when gravity is the only important influence, are said to be in free fall. The direction of this acceleration is downward. This direction is commonly taken to be the negative direction along the axis defining the coordinate system. In this case the acceleration, a is given by a = -g, but notice that the value of g is always positive. In some cases it may be more convenient to choose downward as the positive direction in which case a = +g.

---

Example 2.9 Learning to Juggle: An entertainer is learning to juggle balls thrown very high. One of the balls is thrown vertically upward from 1.80 m above the ground with an initial velocity of 4.92 m/s. If he fails to catch the ball and it hits the ground, how long is it in the air?

Picture the Problem We choose up as the positive direction in this picture. The picture shows the ball's upward trip on the left and downward trip on the right.

Strategy To solve this problem we need an expression that relates time to position, initial velocity, and acceleration. Because of how the coordinate system is chosen,
x₀ = 0, a = -g, and when it hits the ground x = -1.80 m.

Solution

1. Choose the equation relating t with v0, a, and x. Notice that -g has been used for a.
2. Put the quadratic equation for t in standard form.
3. Apply the quadratic formula to get the solutions.
4. The solution that gives a positive time is the correct answer.	t = 0.5015 s + 0.7865 s = 1.29 s

Insight The above solution is only one approach to solving this problem. Another way might be to separately determine the times for the ball to travel upward and downward then add them (try it). The above approach looks at the motion all at once. This can be done because the acceleration is the same going up and coming down. Make sure you can reproduce the numerical values above. Be careful to properly account for all the minus signs. To what does the negative solution correspond?

---

There are a few facts concerning free fall motion that you can utilize in analyzing situations. These facts can be deduced from the four equations for motion with constant acceleration.

• When an object launched vertically upward reaches the top of its path (its maximum height), its instantaneous velocity is zero, even though its acceleration continues to be 9.81 m/s² downward.
  


• An object launched upward from a given height takes an equal amount of time to reach the top of its path as it takes to fall from the top of its path back to the height from which it was launched.
  


• The velocity an object has at a given height, on its way up, is equal and opposite to the velocity it will have at that same height on its way back down.

Physlet Illustration: Acceleration Due to Gravity
Near the surface of another planet, a bouncy ball is dropped from rest. The time is shown in seconds and the velocity in cm/s. Can you determine the acceleration due to gravity on this planet? Start
Hints Choose a specific time interval in the animation, that is, specific starting and stopping times. Read the ball's velocity at the starting time and again at the stopping time. Divide the change in velocity by the time it takes to undergo that change. Does it make any difference whether you study the ball on the way down or the way up? How does gravity on this planet compare to Earth's gravity?

---

Example 2.10 A Child's Toy: A toy rocket is launched vertically upward from the ground. If its initial speed is 8.93 m/s, with what speed will it strike the ground?

Picture the Problem We choose up as the positive direction in this picture. The picture shows the rocket's upward trip on the left and downward trip on the right.

Strategy/Solution Instead of using equations, we can get the solution by knowing the symmetry results above. Since the rocket is launched upward from the ground with a certain speed, we know that when it reaches the ground on its way down it will have the same speed (just in the opposite direction). Therefore, it strikes the ground with a speed of 8.93 m/s.

Insight So sometimes we can use the symmetry to get the answer with little, or no, calculation. Convince yourself of the above result by using the equation to solve the problem.

---

Practice Quiz

A ball is dropped from a height of 2.89 m above the ground. How long does it take to reach the ground? 0.768 s 0.589 s 9.81 s 28.4 s

A ball is thrown vertically upward from a height of 2.00 m above the ground with a speed of 17.3 m/s. If the ball is caught by the same person at the same height, how long is the ball in the air? 1.76 s 4.00 s 3.53 s 8.65 s

Can an object that is moving upward be in free fall? No, because an object cannot be falling if is is moving upward. Yes, because Earth's gravity sometimes pushes objects upward. No, because Earth's gravity never pushes objects upward. Yes, as long as gravity is the only force acting on it.

Stone A is thrown upward from the top of a high bridge while stone B is dropped from the same height. Both stones fall and strike the water below. Which stone strikes the water with greater speed? stone A stone B they strike with equal speeds

Graphical Analysis of Motion

The motion of an object is often analyzed graphically. Graphical analysis is useful for many things and can be used to determine what kinda of motion is being observed. In order to do that we must first know what kinds of graphs the different types of motion produces and how to obtain information from them. Specifically, we focus on graphs of position as a function of time, x-versus-t, and velocity as a function of time,
v-versus-t.

(A) Position-versus-Time

In general, plots of position-versus-time will be curved. Regardless of the shape of the curve however, information about the velocity of the motion can be determined from the graph. Any two points on the graph can be connected by a straight line. The slope of this connecting line equals the average velocity of the object over the corresponding time interval. Also, for any given point on the curve there is a line, called the tangent line, that intersects the curve at that point. The slope of the tangent line at a point equals the instantaneous velocity of the object at the corresponding time.

For the special case of constant velocity motion, the equation x = vt shows that we expect the x-versus-t graph to be linear. The slope of the line would tell us the velocity of the motion.
For the special case of constant acceleration, the equation x = (1/2)at2 (with v0 = 0) shows that we expect the x-versus-t graph to be a parabola. From this curve we can determine average and instantaneous velocities.

(B) Velocity-versus-Time

For motion with constant acceleration, which includes constant velocity motion (a = 0), the graph of velocity as a function of time is linear as can be seen by the equation v = v0 + at. The slope of the straight line equals the acceleration of the motion.
Whether the v-versus-t curve is linear or non-linear (a constant), the distance traveled by an object from one time to another equals the area under the curve between those two times. For the cases of constant velocity and constant acceleration, these areas are rectangles and triangles respectively.

---

Example 2.11 What type of motion is this?: From the following data, determine the type of motion represented. If it is constant velocity motion, find the velocity. If it is motion with constant acceleration, find the acceleration.

t (s):	0	0.50	1.0	1.5	2.0	2.5	3.0
x (m):	0	1.0	2.0	3.0	4.0	5.0	6.0

Picture the Problem For the picture we make a position-versus-time plot of the above data. The data are connected by a dashed line as a visual aid.

Strategy/Solution We examine the features of the plot. It clearly shows a linear relationship so we can conclude that the data is from constant velocity motion. The velocity of the motion is determined by the slope of the line. To calculate the slope we select two widely spaced points on the line (1.0 m, 0.50 s) and (5.0 m, 2.5 s)

Slope equals rise divided by run

Insight Any points along the connecting line would have given the same slope. Real world data points would not fall so neatly on a single straight line. In such cases you would draw a best fit line instead of a connecting line. The velocity would then be measured by the slope of this best-fit line.

---

Physlet Illustration: Constant Acceleration
	Interactive Help on      off Start
The blue ball moves with constant acceleration across the floor, as viewed from above. The distance grid is in cm, and the times are shown in seconds. Can you determine the acceleration of the ball?
Hints At a particular time, find the slope of the distance versus time graph.   How does this slope compare with the velocity at that time?  Note that it is continually increasing. Find the slope of the velocity versus time graph.   What is the acceleration of the ball?

---

Example 2.12 Determine Distance from a Graph: Use the v-versus-t graph below to find the distance traveled by the object whose motion is represented.

Picture the Problem The figure below shows the graph of v-versus-t referred to in the problem.

Strategy To get the distance we need the area under the curve. The diagram shows that this area can be divided into the area of the triangle bounded by the two lines plus the area of the rectangle below this triangle.

Solution

1. Use the formula for the area of a triangle to get one term in the sum x1.
2. Get the area of the rectangle beneath the triangle for the second term in the sum x2.
3. The distance traveled is the sum of the two:	distance = x1 + x2 = 42 m

Insight Notice that the height of the triangle is 12 m/s not 13 m/s. This is because the motion of interest both started and ended with a velocity of 1.0 m/s.

---

Physlet Illustration: Changing Acceleration
	Interactive Help on      off Start
The blue ball moves with a changing acceleration across the floor, as viewed from above. The distance grid is in cm, and the times are shown in seconds. Can you determine the rate of change of the acceleration of the ball?
Hints Find the slope of the distance versus time graph.   How does this slope compare with the velocity?  Note that it is continually increasing. Find the slope of the velocity versus time graph at a couple of times2. Is it Constant?  What is the acceleration of the ball at t = 2.5 seconds?

Practice Quiz

On an x-versus-t plot, the slope of a tangent line equals... average velocity average acceleration instantaneous velocity instantaneous acceleration

If an object's motion is described by a v-versus-t plot that is parabolic in shape, what kinds of motion is it? constant velocity constant acceleration varying acceleration stationary object

Given the x-versus-t graph shown below, which type of motion is most likely represented constant velocity motion accelerated motion motionless particle [none of the above]

prev   2-1 - 2-3   2-4 - 2-6   2-7