Notes 01-03 Page 1 of 8

1-D Motion: Free Falling Objects

So far, we have only looked at objects moving in a horizontal dimension. Today, we’ll look at objects moving in the vertical. Then, we’ll look at both dimensions simultaneously. An object that is moving vertically through the air with no physical constraints on its motion is said to be in free fall. Freely falling objects have a constant acceleration directed toward the center of the Earth that is due only to the Earth’s gravitational field. Free fall can exist on any planet (or moon, etc.), but we tend to focus on Earth. It is important to keep in mind that objects in free fall are always under a constant acceleration. We call the acceleration, “The acceleration due to Earth’s gravity.” We use the letter g to represent it.

𝑇ℎ𝑒 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝐷𝑢𝑒 𝑡𝑜 𝐸𝑎𝑟𝑡ℎ′𝑠 𝐺𝑟𝑎𝑣𝑖𝑡𝑦

�⃗� = 9.80𝑚

𝑠2

This number is valid if you are close to the Earth’s surface (within a few thousand meters).

In reality, this value decreases the farther you get from the Earth’s surface, but we will treat it as a constant! We’ll investigate this thought in more detail later in the course.

Please note that g is a positive number! To better understand free fall, let’s consider a ball that is tossed straight up, reaches a maximum height, and then lands back in the thrower’s hand. We’ll start by analyzing the acceleration over the time of the flight, because it’s constant! If we define “up” as

positive, then we know �⃗� = −9.80𝑚

𝑠2= −�⃗�. The 𝑎 𝑣𝑠. 𝑡 graph would look something like this (I made up some

time values just for the sake of the example):

The acceleration tells us that the v vs. t graph will be linear with a slope of -9.80. This means that the velocity is always becoming more negative (increasing in the “downward” direction). The graph for our example, would look like this:

-12-10

-8-6-4-2024

0 0.5 1 1.5

a (m

/s^2

)

t (s)

-8-6-4-202468

0 0.5 1 1.5v (m

/s)

t (s)

Notes 01-03 Page 2 of 8

Notice that the line passes through zero. This tells us that there is an instant in time where the ball is not moving. This happens at the apex or maximum height of its path! Even though the acceleration is constantly –g, the velocity is still positive for a period of time (while the ball is going up!). Since the object is constantly accelerating, we should expect the y vs. t graph to be parabolic (and it is). The slope of the tangent lines will start off steep and level off to zero as the ball slows down on its way up. The tangent lines will become steeper in the negative direction as the ball speeds up on its way back down. For the graph below, I defined the point of “catch and release” to be a height of zero.

If we could use a camera that took pictures at equal time intervals and overlay the pictures on top of each other, we would see something like this (Obviously I separated the upward and downward paths):

If we always define “up” (away from the Earth) as positive and let y represent the vertical position of an object, then we can rewrite the kinematics equations as follows. Notice, we replace x with y and use 𝑎 = −𝑔.

𝒙 =𝟏

𝟐𝒂𝒕𝟐 + 𝒗𝟎𝒕 + 𝒙𝟎

𝒗 = 𝒂𝒕 + 𝒗𝟎 𝒗𝟐 = 𝒗𝟎

𝟐 + 𝟐𝒂(𝒙 − 𝒙𝟎)

𝒙 = −𝟏

𝟐𝒂𝒕𝟐 + 𝒗𝒕 + 𝒙𝟎

𝚫𝒙 = �̅�𝒕

𝒚 = −𝟏

𝟐𝒈𝒕𝟐 + 𝒗𝟎𝒕 + 𝒚𝟎

𝒗 = −𝒈𝒕 + 𝒗𝟎

𝒗𝟐 = 𝒗𝟎𝟐 − 𝟐𝒈(𝒚 − 𝒚𝟎)

𝒚 =𝟏

𝟐𝒈𝒕𝟐 + 𝒗𝒕 + 𝒚𝟎

𝚫𝒚 = �̅�𝒕

Obviously if we defined “down” (toward the Earth) as positive, then 𝑎 = 𝑔 would be true and we’d have to change our signs in the equations accordingly. NOTE: By assuming g is constant, we are ignoring the effects of air resistance! We have to do this because the mathematics behind air resistance require differential equations (which usually comes after Calculus 3).

0

1

2

3

0 0.5 1 1.5

y(m

)

t (s)

Notes 01-03 Page 3 of 8

Symmetry If the initial and final heights of an object are the same (𝑦 = 𝑦0), then the object rises and falls for an equal amount of time (𝑡𝑟𝑖𝑠𝑒 = 𝑡𝑓𝑎𝑙𝑙). For these conditions to be met, the object must have an initial upward velocity. To

see the symmetry of this situation graphically, look at the y vs. t graph on the previous page. Problems 1. In a moment of rage, you throw your physics book as high into the air as you can. What is the book’s velocity when the book is at its highest possible point? What is the book’s acceleration at this point?

At the highest point, 𝒗 = 𝟎 𝒂𝒏𝒅 𝒂 = −𝒈.

2. At the end of the last Apollo 15 moon walk, Commander David Scott performed a live demonstration for the

television cameras. He held out a geologic hammer and a feather and dropped them from the same height at the same time. Which object hit the ground first? (Note: There is virtually no atmosphere on the moon – the moon exists in a near-vacuum.) Since there is no appreciable air resistance, the hammer and feather will experience the same acceleration – that due to the moon’s gravitational field. Each object has a starting velocity of zero and fall the same height. Thus, the hammer and the feather will strike the ground at the same time. There is a video of this on YouTube 3. A tennis player on serve tosses the ball straight up before hitting it. What happens to the speed of the ball as it travels through the air? The speed of the ball will decrease until the ball reaches its highest position. The speed will then increase as the ball begins falling back downward. (Note to me: Sketch a kinematics graph of this situation).

4. In a fit of rebellion, two AP Physics students drop their physics books from the top of a tall building. Student A drops his book first and then Student B drops her book. Describe what happens to the distance between the two books while they are both still falling. (Hint: If you are struggling with this problem, create a table of some mock data for the velocity of each book at some various times.)

When student B drops her book, Student A’s book already has a downward velocity. Since the two books accelerate at the same rate, Student A’s book will always have a greater velocity than student B’s book. This means that the distance between the two books will continually increase until Student A’s book strikes the ground. This situation is comparable to having two identical cars in a drag race. If one car gets a head start and the two cars have identical accelerations, then the first car will always be moving faster and thus the distance between the two cars will keep increasing.

Notes 01-03 Page 4 of 8

5. In our pre-lab discussion for Lab 2, we dropped a ball from the same height of a person’s head and timed how long it took for the ball to hit the ground. We then used the equation ℎ = 4.9𝑡2 to estimate the person’s height. (a) Show why this equation is valid for this scenario. (b) Suppose it took a ball 0.55 seconds to fall to the ground when dropped from the same height of a certain person. How tall is the person? (c) Mr. Scheithauer is approximately 1.88 m tall. If a ball was dropped from his height, how long would it take to strike the ground?

(a) 𝒚 = −𝟏

𝟐𝒈𝒕𝟐 + 𝒗𝟎𝒕 + 𝒚𝟎. We let 𝒚 = 𝟎 (ground), 𝒗𝟎 = 𝟎 (dropped), and 𝒚𝟎 = 𝒉 (person’s height).

This yields, 𝟎 = −𝟏

𝟐𝒈𝒕𝟐 + 𝒉 ⇒ 𝒉 =

𝟏

𝟐𝒈𝒕𝟐 =

𝟏

𝟐(𝟗. 𝟖

𝒎

𝒔𝟐) 𝒕𝟐 = (𝟒. 𝟗

𝒎

𝒔𝟐) 𝒕𝟐.

(b) 𝒉 = (𝟒. 𝟗𝒎

𝒔𝟐) (𝟎. 𝟓𝟓 𝒔)𝟐 = 𝟏. 𝟒𝟖 𝒎

(c) 𝒉 = 𝟒. 𝟗𝒕𝟐 ⇒ 𝒕 = √𝒉

𝟒. 𝟗𝒎𝒔𝟐

= √𝟏.𝟖𝟖 𝒎

𝟒. 𝟗𝒎𝒔𝟐

= 𝟎. 𝟔𝟐 𝒔

6. An arrow is shot straight upward at 100 m/s. (a) If air resistance is neglected, what is the maximum height the arrow would reach? (b) How long would the arrow be in the air? (c) Suppose the arrow was shot on the surface of Mars instead. How would the values for parts (a) and (b) compare on Mars to those calculated on Earth? (The

acceleration due to gravity at the surface of Mars is approximately 3.7𝑚

𝑠2.

(𝒂) 𝒗𝟐 = −𝟐𝒈𝚫𝒚+ 𝒗𝟎𝟐

𝚫𝒚 = −𝒗𝟎𝟐

𝟐𝒈= −

(𝟏𝟎𝟎𝒎𝒔)𝟐

𝟐(𝟗. 𝟖𝟎𝒎𝒔𝟐)= 𝟓𝟏𝟎. 𝟐 𝒎/𝒔

(b) 𝑵𝒐𝒕𝒆:𝒗 = −𝒗𝟎

𝒗 = −𝒈𝒕 + 𝒗𝟎

𝒕 =𝟐𝒗𝟎𝒈=𝟐(𝟏𝟎𝟎

𝒎𝒔 )

𝟗. 𝟖𝟎𝒎𝒔𝟐

= 𝟐𝟎. 𝟒 𝒔

(c) We could do the calculations, but this isn’t necessary. Looking at 𝚫𝐲 = −𝐯𝟎𝟐

𝟐𝐠 tells us that as g decreases, 𝚫𝐲

increases. Further, looking at 𝐭 =𝟐𝐯𝟎

𝐠 shows that as g decreases, t increases. Such analysis shows the benefits of

performing algebraic manipulation before substituting in numerical values.

Notes 01-03 Page 5 of 8

7. A sky diver is using a camera to film his jump. Near the end of his jump, when he is at a height of 50m and falling at a constant rate of 10 m/s, he accidentally drops his camera. (a) With what velocity does the camera strike the ground? (b) How many seconds does it take for the camera to strike the ground after it is dropped? (c) What if (for some bizarre reason) the sky diver threw the camera upward with an upward velocity of 10m/s (relative to the ground) and let the camera fall to the ground. How would the velocity that the camera struck the ground with compare to that of part (a)? How would the time of flight compare to that of part (b)?

(𝒂) 𝒗𝟐 = −𝟐𝒈𝚫𝒚+ 𝒗𝟎𝟐

𝒗 = −√−𝟐(𝟗. 𝟖𝟎𝒎

𝒔𝟐) (−𝟓𝟎 𝒎) + (−𝟏𝟎

𝒎

𝒔)𝟐

⏟ Explain Negatives

= −𝟑𝟐. 𝟖𝟔 𝒎/𝒔

(𝒃) 𝒗 = −𝒈𝒕 + 𝒗𝟎

𝒕 =𝒗 − 𝒗𝟎−𝒈

=(−𝟑𝟐. 𝟖𝟔

𝒎𝒔− (−𝟏𝟎

𝒎𝒔))

−𝟗. 𝟖𝟎𝒎𝒔𝟐

= 𝟐. 𝟑𝟑 𝒔

(c) The velocity would be the same. The camera would initially rise and return to the height it was originally

thrown from. At this point it will have a downward velocity of 10m/s, which is the exact scenario described in

part a. The time of flight would obviously be larger.

8. A ball is thrown vertically upward with a speed of 25.0 m/s, rises to a maximum height, and then falls, eventually striking the ground.. (a) How high does the ball rise from its point of release? (b) How many seconds does it take for the ball to reach its highest position? (c) How many seconds does it take for the ball to strike the ground after it has reached its maximum height? (d) What is the ball’s velocity when it returns to the same height from which it was released?

Notes 01-03 Page 6 of 8

9. Two students are on a balcony that is a height ℎ above the ground. They each throw a ball with an initial speed of 𝑣0; one throws the ball upward and the other downward. (a) With what speed does each ball strike the ground? (b) Find the difference in the “flight times” of each ball. (c) Find an equation that describes the distance between the balls while they are both in the air. Answer each question in terms of 𝑣0, 𝑔, ℎ, 𝑎𝑛𝑑 𝑡.

(𝒃) 𝒕𝟏 =𝒗 − 𝒗𝟎−𝒈

=

(−√𝟐𝒈𝒉+ 𝒗𝟎𝟐 − (−𝒗𝟎))

−𝒈=√𝟐𝒈𝒉 + 𝒗𝟎

𝟐

𝒈−𝒗𝟎𝒈

Similarly…

𝒕𝟐 =√𝟐𝒈𝒉+ 𝒗𝟎

𝟐

𝒈+𝒗𝟎𝒈

𝒕𝟐 − 𝒕𝟏 =√𝟐𝒈𝒉 + 𝒗𝟎

𝟐

𝒈+𝒗𝟎𝒈−

(

√𝟐𝒈𝒉+ 𝒗𝟎

𝟐

𝒈−𝒗𝟎𝒈

)

=𝟐𝒗𝟎𝒈

(c)

Notes 01-03 Page 7 of 8

10. A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates upward at 2.00𝑚

𝑠2

until it reaches an altitude of 150 m. (a) What is the maximum height reached by the rocket? (b) How long does it

take for the rocket to reach this maximum height? (c) How long is the rocket in the air for?

Notes 01-03 Page 8 of 8

11. You are bored so you decide to execute a simple physics experiment. A friend drops an object from above a 1.5m-tall window in your house. Through video analysis, you find that the object travels from the top of the window to the bottom of the window in 0.14 seconds. How high above the top of the window was the object dropped? *Draw a figure on the board Using “down” as the positive direction:

𝚫𝒚 =𝟏

𝟐𝒈𝒕𝟐 + 𝒗𝒕𝒐𝒑𝒕

𝒗𝒕𝒐𝒑 =(𝚫𝒚 +

𝟏𝟐𝒈𝒕𝟐)

𝒕

𝒗𝒕𝒐𝒑 =

((𝟏. 𝟓𝟎 𝒎) + (𝟏𝟐) (

𝟗. 𝟖𝟎𝒎𝒔𝟐) (𝟎. 𝟏𝟒 𝒔)𝟐)

𝟎. 𝟏𝟒 𝒔= 𝟏𝟏. 𝟒𝟎 𝒎/𝒔

𝒗𝒊𝒏𝒊𝒕𝒊𝒂𝒍𝟐 = −𝟐𝒈𝒉 + 𝒗𝒕𝒐𝒑

𝟐

𝒉 =𝒗𝒕𝒐𝒑𝟐

𝟐𝒈= 𝟔. 𝟔𝟑 𝒎

⏟ 𝒉 𝒂𝒃𝒐𝒗𝒆 𝒕𝒐𝒑 𝒐𝒇 𝒘𝒊𝒏𝒅𝒐𝒘