
The term free fall describes any state of motion in which an object has a constant downward acceleration due only to gravity. Near Earth’s surface, the magnitude of that acceleration is 9.8 m/s^{2} (also known as one “g”). An object can be in free fall even if it is rising. When a rising projectile reaches its maximum height, its vertical velocity is momentarily zero (i.e., it is neither rising nor falling). A falling object reaches terminal velocity when the upward force of air resistance on the object equals the object’s weight.

free fall, terminal velocity

Review problems and questions 

 Which of the following objects are in free fall? (Assume that you can ignore air resistance.)
 a textbook lying on a desk
 a textbook that has been tossed upward and has left the student’s hands but not yet touched any other object or surface
 a textbook that has been thrown downward from a tall cabinet
 a textbook held tightly by a running student

 No. This textbook is not in free fall because it is not accelerating, and because a force from the table is acting on it in addition to the force of gravity.
 Yes. Although this object is rising, not falling, it is in free fall because gravity is the only force acting on it.
 Yes. Once the textbook is released, it is in freefall because gravity is the only force acting on it. The force that gave it the initial downward velocity is no longer being applied.
 No. The book may be accelerating, but it is not accelerating due only to the force of gravity acting upon it.

 Josh tells Amanda that acceleration must be a negative number whenever an object is in free fall. “After all,” he claims, “gravity pulls objects downward, and down is negative.” Amanda replies that the acceleration in a free fall problem can be negative or positive. “You just have to make sure all related quantities, like displacement and velocity, are handled the same way,” she adds. Who is right, and why? (Answer in complete sentences, and cite a specific passage in the text to support your claim.)

Amanda is correct. According to “The sign of g” (page 228), the acceleration due to gravity can be given either a positive sign or a negative sign. However, once we choose a sign convention for g, we must take care to be consistent and treat other vectors the same way. (For example, if we allow g to be positive, we must treat downward displacements and velocities as positive as well.)

 Zorgata drops her ball a distance of 1 m on the surface of planet Zogg. Izqif wants his ball to take twice as much time to fall as Zorgata’s. What height should he drop it from? (Note that gravity is stronger on the surface of Zogg than it is on Earth.)
 2 m
 3 m
 4 m
 You cannot say without knowing the value for g on planet Zogg.

Choice c is correct. Izqif should drop his ball from a height of 4 m. The reason is that distance is proportional to the square of time in the formula d = (x−x_{0}) = ½at^{2}, which applies when objects are dropped rather than thrown (because the initial velocity, v_{0}, is zero). For Izqif’s ball to take twice as long as Zorgata’s, Izqif must raise his ball four times as high before releasing it. This statement is true no matter how strong the acceleration due to gravity is.

 During the Apollo 15 mission, astronaut David Scott performed a scientific experiment on the surface of the Moon. Scott dropped two objects at the same time from the same height. One was a 1.3 kg hammer; the other was a 0.03 kg feather. He dropped the two objects from a height of 1.6 m. The two objects fell, and both hit the Moon’s surface 1.4 s after being released.
 What is the acceleration due to gravity on the Moon’s surface, in units of meters per second squared?
 Were Scott’s hammer and feather both in free fall?

Answer: 1.6 m/s^{2}
 Yes; both objects were in free fall.
Solution: 1.6 m/s^{2}. We begin by assuming that the hammer is in free fall and that a equals the acceleration due to gravity.
Asked: acceleration of hammer a Given: Falling distance d = 1.6 m; falling time t = 1.4 s; initial velocity v_{0} = 0 Relationship: d = ½ at^{2} Solution: First note that the distance fallen equals the difference (d) between the final and initial positions (x−x_{0}): $$d={v}_{0}t+\frac{1}{2}a{t}^{2}$$ Next, note that the initial velocity is zero, since Scott dropped the hammer, rather than throwing it upward or downward, so $$d=\frac{1}{2}a{t}^{2}$$ Now insert the given values for the distance and time interval: $$1.6\text{m}=\frac{1}{2}a{\left(1.4\text{s}\right)}^{2}$$ Rearrange numerical quantities and divide both sides by m^{2} to get the SI unit of acceleration: $$1.6\frac{\text{m}}{{\text{s}}^{\text{2}}}=\frac{{1.4}^{2}}{2}a$$ Rearrange so that the unknown quantity (a) is on the left side of the equation and everything else is on the right: $$a=\frac{2(1.6)}{{1.4}^{2}}\frac{\text{m}}{{\text{s}}^{\text{2}}}\cong 1.6\frac{\text{m}}{{\text{s}}^{\text{2}}}$$ The hammer’s acceleration was 1.6 m/s^{2} downward—about onesixth of the “strength” of gravity on Earth.  Both objects were in free fall.

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