Falling Objects
By the end of this section, you will be able to:
Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.
The most remarkable and unexpected fact about falling objects is that, if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass. This experimentally determined fact is unexpected, because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones.
Figure 1 A hammer and a feather will fall with the same constant acceleration if air resistance is considered negligible. This is a general characteristic of gravity not unique to Earth, as astronaut David R. Scott demonstrated on the Moon in 1971, where the acceleration due to gravity is only 1.67 \(\mathrm{m/s}^2\).
In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball will reach the ground after a hard baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, while friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them. For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall.
The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called the acceleration due to gravity. The acceleration due to gravity is constant, which means we can apply the kinematics equations to any falling object where air resistance and friction are negligible. This opens a broad class of interesting situations to us. The acceleration due to gravity is so important that its magnitude is given its own symbol, \(g\). It is constant at any given location on Earth and has the average value.
Although g varies from 9.78 \(\mathrm{m/s}^2\) to 9.83 \(\mathrm{m/s}^2\), depending on latitude, altitude, underlying geological formations, and local topography, the average value of 9.80 \(\mathrm{m/s}^2\) will be used in this text unless otherwise specified. The direction of the acceleration due to gravity is downward (towards the center of Earth). In fact, its direction defines what we call vertical. Note that whether the acceleration a in the kinematic equations has the value +g or −g depends on how we define our coordinate system. If we define the upward direction as positive, then \(a=−g=−9.80\) \(\mathrm{m/s}^2\), and if we define the downward direction as positive, then \(a=g=9.80\) \(\mathrm{m/s}^2\).
The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So we start by considering straight up and down motion with no air resistance or friction. These assumptions mean that the velocity (if there is any) is vertical. If the object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude \(g\). We will also represent vertical displacement with the symbol \(y\) and use \(x\) for horizontal displacement.
What will happen when a coin and a feather are simultaneously dropped from the same height? (Neglect air resistance)
Which two of the statements describe an object in free fall?
A nail and a cotton are dropped at the same time from rest in a vacuum. The nail reaches the surface/ground ______________:
The acceleration of free-falling objects is called:
At the very instant at the same altitude, a heavy person and a light person open their parachute of the same size, together. Who would reach the ground first?
Kinematic Equations
A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s. The rock misses the edge of the cliff as it falls back to Earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance.
Strategy
Draw a sketch.
Figure 1
We are asked to determine the position \(y\) at various times. It is reasonable to take the initial position \(y_0\) to be zero. This problem involves one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, so \(a\) is negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion and will slow and eventually reverse it.
Since we are asked for values of position and velocity at three times, we will refer to these as \(y_1\) and \(v_1\); \(y_2\) and \(v_2\); and \(y_3\) and \(v_3\).
Solution for Position \(y_1\)
1. Identify the knowns. We know that \(y_0\ =\ 0 ;\ v_0\ =\ 13.0 \;m/s; \ a\ =\ −g=−9.80 \;m/s^2;\ and\ t=1.00 \;s\)
2. Identify the best equation to use. We will use \(y=y_0+v_0t+\frac{1}{2}at^2\) because it includes only one unknown, \(y\) (or \(y_1\)here), which is the value we want to find.
3. Plug in the known values and solve for \(y_1\).
Discussion
The rock is 8.10 m above its starting point at \(t=1.00\;s\), since \(y_1>y_0\). It could be moving up or down; the only way to tell is to calculate \(v_1\) and find out if it is positive or negative.
Solution for Velocity \(v_1\)
1. Identify the knowns. We know that \(y0=0 ;\ v0=13.0\; m/s;\ a=−g=−9.80 \;m/s^2;\ and \ t=1.00 \;s.\) We also know from the solution above that \(y_1=8.10\;m\).
2. Identify the best equation to use. The most straightforward is \(v=v_0−gt\) (from \(v=v_0+at\), where \(a=gravitational\ acceleration=−g\)).
3. Plug in the knowns and solve.
The positive value for \(v_1\) means that the rock is still heading upward at \(t=1.00\;s\). However, it has slowed from its original 13.0 m/s, as expected.
Solution for Remaining Times
The procedures for calculating the position and velocity at \(t=2.00s\) and \(3.00\) are the same as those above. The results are summarized in Table 1 and illustrated in Figure 2.
Graphing the data helps us understand it more clearly.
Figure 2 Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that velocity changes linearly with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that the graph shows some horizontal motion—the shape of the graph looks like the path of a projectile. But this is not the case; the horizontal axis is time, not space. The actual path of the rock in space is straight up, and straight down.
The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since y1 and v1 are both positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both \(y_3\) and \(v_3\) are negative, meaning the rock is below its starting point and continuing to move downward. Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its acceleration is still \(−9.80 \;m/s^2\). Its acceleration is \(−9.80\, m/s^2\) for the whole trip—while it is moving up and while it is moving down. Note that the values for \(y\) are the positions (or displacements) of the rock, not the total distances traveled. Finally, note that free-fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while arcing up as well as down, as we will discuss in more detail later.
A soft tennis ball is dropped onto a hard floor from a height of 2.32 m and rebounds to a height of 1.41 m. Calculate its velocity, in meters per second, just before it strikes the floor.
Calculate its velocity just after it leaves the floor on its way back up.
Express your answer in meters per second.
Calculate its acceleration during contact with the floor if that contact lasts 3.4 ms (0.0034 s).
Calculate the displacement at \(t= 0.6 \;s\) for a ball thrown straight up with an initial velocity of \(18\;{m \over s}\). Take the point of release to be \(y_0 = 0\).
Express your answer in meters.
Calculate the displacement, in meters, at \(t= 1.3 \;s\).
Calculate the velocity, in meters per second, at \(t= 0.6 \;s\).
Calculate the velocity, in meters per second, at \(t= 1.3 \;s\).
Calculate the displacement at 0.5 s for a rock thrown straight down with an initial velocity of \(19\;{m \over s}\) from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
Calculate the displacement in meters at 2.6 s.
Calculate the velocity, in meters per second, at 0.5 s.
Calculate the velocity at 2.6 s.
A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.38 m above the floor in an attempt to get the ball?
A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of \(1.5\;{m \over s}\) and observes that it takes 1.7 s to reach the water. How high, in meters, above the water was the preserver released?
Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver so that an acceleration equal to that of gravity is reasonable.
Calculate the height, in meters, of a cliff if it takes 2.48 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of \(7\;{m \over s}\).
How long would it take to reach the ground if it is thrown straight down with the same speed?
Express your answer in seconds.
A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of \(13.3\;{m \over s}\). How long does he have to get out of the way if the shot was released at a height of 2.35 m, and he is 2.06 m tall?
Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break lose from a height of 129 m. He can’t see the rock right away but sees it 1.74 s later. How far above the hiker is the rock when he can see it?
How much time does he have to move before the rock hits his head?
An object is dropped from a height of 59 m above ground level. Determine the distance in meters traveled during the first 2 seconds.
Determine the final velocity, in meters per second, at which the object hits the ground.
Determine the distance traveled during the last second of motion before hitting the ground.
A coin is dropped from a hot-air balloon that is 268 m above the ground and rising at \(13\;{m \over s}\) upward. For the coin, find the maximum height reached.
Calculate its velocity 4 s after being released. Express your answer in meters per second.
Calculate its position, in meters, 4 s after being released.
A stone is thrown into a pit and takes 3 seconds to reach the bottom. How deep is the pit?
A parachutist jumps from a plane and falls free for 37 seconds until he opens his parachute 117 meters from the ground. How high was the plane flying when the parachutist jumped?
What is the displacement of a free-falling body after 13 seconds if it starts from rest? Neglect air resistance and express your answer in meters.
Calculate the final velocity, in meters per second, of a stone to reach a height of 2 m by throwing it vertically upward.
A girl on a bridge throws a ball vertically downwards with an initial velocity of 14 \(m \over s\). If the ball reaches the floor 1 s later, how high in meters is the bridge?
Reaction Time
A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between your thumb and index finger, separated by about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler unexpectedly, and try to catch it between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity, calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot to go from the gas pedal to the brake was twice this reaction time?
What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question, calculate the velocity of the rock when it is 5.10 m below the starting point, and has been thrown downward with an initial speed of 13.0 m/s.
Since up is positive, the final position of the rock will be negative because it finishes below the starting point at \(y_{0}=0.\) Similarly, the initial velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the rock will continue to move downward.
Solution
1. Identify the knowns. \(y_0=0;\ y_1=−5.10 m; \ v_0=−13.0 m/s; \ a=−g=−9.80 m/s^2.\)
2. Choose the kinematic equation that makes it easiest to solve the problem. The equation \(v^2=v_0^2-2g(y−y0)\) works well because the only unknown in it is \(v\). (We will plug \(y_1\) in for \(y\).)
3. Enter the known values
where we have retained extra significant figures because this is an intermediate result. Taking the square root, and noting that a square root can be positive or negative, gives
The negative root is chosen to indicate that the rock is still heading down. Thus,
Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed. (See Example 1 and Figure 2(a).) This is not a coincidental result. Because we only consider the acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical position relative to the starting point. For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example 1) when the initial velocity is 13.0 m/s straight up, a result of \(\pm3.20\; m/s\) is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction.
Another way to look at it is this: In Example 1, the rock is thrown up with an initial velocity of 13.0 \(\mathrm{m/s}\). It rises and then falls back down. When its position is \(y=0\) on its way back down, its velocity is −13.0 \(\mathrm{m/s}\). That is, it has the same speed on its way down as on its way up. We would then expect its velocity at a position of \(y=−5.10 \thinspace \mathrm{m}\) to be the same whether we have thrown it upwards at +13.0 \(\mathrm{m/s}\) or thrown it downwards at −13.0 \(\mathrm{m/s}\). The velocity of the rock on its way down from \(y=0\) is the same whether we have thrown it up or down to start with, as long as the speed with which it was initially thrown is the same.
The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics laboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, Figure 3. Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.
Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location?
We need to solve for acceleration \(a\). Note that in this case, displacement is downward and therefore negative, as is acceleration.
1. Identify the knowns. \(y_{0}=0; y=–1.0000 \thinspace \mathrm{m}\); \(t=0.45173\); \(v_{0}=0\).
2. Choose the equation that allows you to solve for \(a\) using the known values.
3. Substitute 0 for \(v_{0}\) and rearrange the equation to solve for \(a\). Substituting 0 for \(v_{0}\) yields
Solving for \(a\) gives
4. Substitute known values yields
so, because \(a=-g\) with the directions we have chosen,
The negative value for \(a\) indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of 9.80 \(\mathrm{m/s^2}\), so 9.8010 \(\mathrm{m/s^2}\) makes sense. Since the data going into the calculation are relatively precise, this value for \(g\) is more precise than the average value of 9.80 \(\mathrm{m/s^2}\); it represents the local value for the acceleration due to gravity.(-)
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of \(6\;{m \over s}\), and her takeoff point is 0.81 m above the pool. How long are her feet in the air?
What is her highest point above the board (in meters)?
What is her velocity in meters per second when her feet hit the water?
A dolphin in an aquatic show jumps straight up out of the water at a velocity of \(13.4\;{m \over s}\). How high, in meters, does its body rise above the water?
How long is the dolphin in the air? Neglect any effects due to his size or orientation and express your answer in seconds.
You throw a ball straight up with an initial velocity of \(15.1\;{m \over s}\). It passes a tree branch on the way up at a height of 2.26 m. How much additional time will pass before the ball passes the tree branch on the way back down?
A kangaroo can jump over an object 3 m high. Calculate its vertical speed, in meters per second, when it leaves the ground.
How many seconds is it in the air?
A 131-g stone is dropped from the top of a 109-m building. How many seconds will it take for the stone to reach the ground?
A steel ball is dropped onto a hard floor from a height of 2.8 m and rebounds to a height of 2.19 m. Calculate its velocity, in meters per second, just before it strikes the floor.
Calculate its velocity, in meters per second, just after it leaves the floor on its way back up.
Calculate its acceleration during contact with the floor if that contact lasts 0.06 ms (1E-04 s).
How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
A bullet was shot at 574 \(m \over s\) and traveled 274 m to the target. Calculate the time (in seconds) it will take for the bullet to reach the target.
Calculate the deviation of the bullet due to gravitational attraction. Express your answer in meters.
A rock is thrown vertically upward with a velocity of 16 \(m \over s\) from the edge of a bridge 37 m above a river. What is the rock's speed (in meters per second) just before it falls into the river?
How many seconds does it take from the time the rock is launched to the time when the rock strikes the river water.
There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. How fast will it be going when it strikes the ground?
Assuming a reaction time of 0.23 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.
A ball is dropped from a window 11 m above the sidewalk. Determine the time it takes for the ball to fall to the sidewalk.
A camera is accidentally dropped from the edge of a cliff and 9 s later hits the bottom. How fast, in meters per second, has just before hitting?
How high in meters is the cliff?
Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.75 s.
Science Practices
While it is well established that the acceleration due to gravity is quite nearly 9.8 \(\mathrm{m/s}^2\) at all locations on Earth, you can verify this for yourself with some basic materials. Your task is to find the acceleration due to gravity at your location. Achieving an acceleration of precisely 9.8 \(\mathrm{m/s}^2\) will be difficult.
However, with good preparation and attention to detail, you should be able to get close. Before you begin working, consider the following questions. Your task is to find the acceleration due to gravity at your location. Achieving an acceleration of precisely 9.8 \(\mathrm{m/s}^2\) will be difficult. However, with good preparation and attention to detail, you should be able to get close. Before you begin working, consider the following questions.
Upon completing these four questions, record your procedure. Once recorded, you may carry out the experiment. If you find that your experiment cannot be carried out, you may revise your procedure.
Once you have found your experimental acceleration, compare it to the assumed value of 9.8 \(\mathrm{m/s}^2\). If error exists, what were the likely sources of this error? How could you change your procedure in order to improve the accuracy of your findings?
Your task is to find the acceleration due to gravity at your location. Achieving an acceleration of precisely 9.8 \(\mathrm{m/s}^2\) will be difficult. However, with good preparation and attention to detail, you should be able to get close. Before you begin working, consider the following questions.
Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. \(y =bx\)) to see how they add to generate the polynomial curve.
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