Free fall refers to the motion of objects influenced only by gravity. The acceleration due to gravity, g, is different for different planets and at different heights above the planet. For motion near the Earth's surface (between sea level and the top of Mount Everest) g = 9.8 m/s^{2}. (Related search term: free falling body problems; freefalling)
EXAMPLE
1.
a. How long does it take a ball to fall from a roof to the ground 7.0 m below?
b. With what speed does it strike the ground?
ANSWER:
In kinematics problems, start off with a dav_{f}v_{i}t table. Use this format to list the information given and identify the quantity being solved for. Then identify the relationship between the given and the unknown quantities, substitute the values into the relationship, and solve fro the unknown.
1.
d 
7.0 m [down] 
a 
9.8 m/s^{2} [down] 
v_{f} 

v_{i} 
0 
t 
? 
a. Now we can see we need a relationship between d, a, v_{i}, and t
d = v_{i}t + (0.5)at^{2}
(7.0 m) = (0)t + (0.5)(9.8)t^{2}
t = 1.20 s
b. We need a relationship between d, a, v_{i}, and v_{f}
v_{f}^{2} = v_{i}^{2} + 2ad
v_{f}^{2} = 0^{2} + 2(9.8)(7.0)
v_{f} = 11.7 m/s
THROWUP PROBLEMS
Throw up problems refer to situations where an object's initial velocity is opposite to its acceleration. The key is to choose a frame of reference. For example, if "up" is +, then "down" is . The frame of reference must be used consistently throughout the solving process.
EXAMPLE
2. How long does it take a ball to reach the ground 7.0 m below, if it is thrown straight up with an initial speed of 2.00 m/s?
ANSWER:
2. frame of reference: down = +
d 
7.0 m 
a 
9.8 m/s^{2} 
v_{f} 

v_{i} 
 2.0 m/s * 
t 
? 
* Do not use "[up]" and "" at the same time unless you are trying to say "the opposite of [up]")
Now we can see we need a relationship between d, a, v_{i}, and t
d = v_{i}t + (0.5)at^{2}
(7.0 m) = (2.00)t + (0.5)(9.8)t^{2}
t = {1.42, 1.01}
Since t < 0 has no meaning,
t = 1.42 s
CATCHUP PROBLEMS
In catchup problems, two objects with different motions end up at the same place at the same time. Sometimes, these problems have the appearance of not having enough information to be solved. However, in physics we trust.
These problems are complex because they describe two different motions. The approach used is to simplify the problem by breaking it down into simple problems. This is down by using two columns in the dav_{f}v_{i}t table: one column for each motion.
EXAMPLE
3. A ball is dropped from a roof to the ground 8.0 m below. A rock is thrown down from the roof 0.600 s later. If they both hit the ground at the same time, what was the initial speed of the rock?
ANSWER:
3.

ball 
rock 
d 
8.0 m [down] 
8.0 m [down] 
a 
9.8 m/s^{2} [down] 
9.8 m/s^{2} [down] 
v_{f} 


v_{i} 
0 
? 
t 
? 
? 
We need time to find speed. Since we have more information about the ball we start off solving for time for the ball.
Now we can see we need a relationship between d, a, v_{i}, and t
d = v_{i}t + (0.5)at^{2}
8.0 = (0)t + (0.5)(9.8 m/s2)t^{2}
t = 1.28 s
The time of travel for the rock is 0.600 s less than for the ball. Now our table looks like this:

ball 
rock 
d 
8.0 m [down] 
8.0 m [down] 
a 
9.8 m/s^{2} [down] 
9.8 m/s^{2 }[down] 
v_{f} 


v_{i} 
0 
? 
t 
1.28 s 
0.68 s 
For the rock we need a relationship between d, a, v_{i}, and t
d = v_{i}t + (0.5)at^{2}
8.0 = v_{i}(0.68 s) + (0.5)(9.8 m/s2)(0.68 s)^{2}
v_{i} = 8.43 m/s
TERMINAL VELOCITY
EXAMPLE
4. Describe the motion of a parachute.
ANSWER:
4. We presume the question is to describe the motion of an object attached to a parachute. Although parachutes have other uses, the discussion will be restricted to the motion of a falling payload, as opposed to the motion of a dragster, or aerobatic skydiver. In the most simplified case, the object continues to fall straight down.
In a vacuum, or on a moon or planet with no atmosphere, a falling object will continue to accelerate until it hits the ground. However, in air (or any other fluid) drag is created as the object falls through the air. Drag is friction between the moving object and the surrounding air which resists the motion of the object.
Drag increases as the square of the velocity of the object, and also depends on the viscosity of the surrounding fluid. So when the speed of the object is 3 times greater, the drag on the object is 9 times greater. Also, there is more drag near the ground where the atmosphere is denser than at high altitudes where the air is “thinner.” As an object falls through the atmosphere picking up speed, drag becomes great enough that the object stops accelerating and continues falling at a constant speed. The maximum speed at which an object will fall is called “terminal velocity.”
Drag also depends on the mass and surface area of the falling object. A feather takes longer to reach the ground than a much larger text book because of its ratio of surface area to mass. The larger the area and the less the mass, the lower the terminal velocity is. The main function of a parachute is to create drag, although some parachutes are designed to create lift. When the canopy opens, the effect is to increase the surface area of the falling object.
The actual motion created by a parachute depends on the design of the parachute and the speed of the object when the parachute opens. Parachutes can be designed to open slowly so that drag is increased slowly at a comfortable rate, or they can open more quickly which has a sudden pronounced effect and can cause strain or discomfort. One possibility is the object has not reached terminal velocity when the parachute opens. In this case the object will continue to pick up speed, but the downward acceleration of the falling object decreases to zero. At this point the object falls with constant speed. Another possibility is that the object has reached or exceeded terminal velocity with the parachute open. In this case the object will decelerate, continuing to fall at a slowing rate until terminal velocity is reached. At this point the object will fall with constant speed.
A. Drop /Throw Down Problems
(See below for answers)
1. A peso is dropped into a well and it falls for 3 seconds before hitting the water. What is the peso's average speed during its 3 second drop?
2. A rock is dropped from the top of an overhang and strikes the ground 6.5 seconds later. How high is the overhang in meters?
3. It takes 0.210s for a dropped wrench to travel past a poster that is 1.35 meters tall. How high above the top of the poster was the wrench released?
4. A falling paratrooper (100 kg with parachute) experiences air resistance equal to 25% of his weight. What is his acceleration?
5. Consider a transparent elevator accelerating upward with acceleration equal to that of the gravity. If a rock was dropped inside the elevator, what would an observer on the ground see the rock do?
B. Throw Up Problems
(See below for answers)
1. You throw a ball downward from a window at a speed of 2.0 m/s. The ball accelerates at 9.8 m/s^{2}.
a. How fast is it moving when it hits the sidewalk 2.5 m below?
b. If you throw the same ball up instead of down, how fast is it moving when it hits the sidewalk?
2. A ball is thrown straight up with a speed of 4.6 m/s. How long does the ball take to reach its maximum height?
3. A round is launched straight up at 460 m/s. How long will it take it to reach its apex and how high will that be? (air resistance may be neglected.)
4. What height will a dart achieve 7 seconds after being blown straight up at 50 m/s?
5. An apple thrown straight upward rises to 24 m above its launch point. At what height has apple's speed decreased to onehalf of its initial value?
6. A stone is flung straight up from a point 1.50 m above the ground and with an initial speed of 19.6 m/s.
a. What is the stone's maximum height above the ground?
b. How much time passes before the stone hits the ground?
7. A blimp is hovering above the ground. When the pilot drops a sandbag overboard, the blimp rises with a constant velocity of 2 m/s. At the moment the sandbag hits the ground, the blimp is 50 m above the ground.
a. How far above the ground was the blimp when the sandbag was dropped?
b. When the sandbag is halfway to the ground, what is its acceleration?
8. A helicopter is ascending vertically with a speed of 5.00 m/s. At a height of 105 m above the ground, a package is dropped from a window. How much time does it take for the package to reach the ground?
9. A rock is thrown up at an initial velocity of 9.8 m/s. What is the time it takes to hit the ground?
C. Catch Up Problems
(See below for answers)
1. An archer shoots an arrow with an initial velocity of 21 m/s straight up from his bow. He quickly reloads and shoots another arrow in the same way 3.0 s later. At what time and height do the arrows meet?
2. A hoist is lifting a naturalist to the top of a cliff at 2.03 m/s vertically. The naturalist suddenly realizes she has left her pet rock behind. A friend picks it up and tosses it straight upward. If the naturalist is 2.50 m above her friend, what is the minimum initial speed the pet rock must have to reach the naturalist?
3. A boy shoots a rock from his slingshot at a target just as the target drops from a tree branch. Should the boy aim a little above the target, a little below the target, or straight at the target?
D. Other Problems
(See below for answers)
1. What is drag?
2. What is the percent difference between the acceleration due to gravity at sea level and at the uppermost peak of Mount Everest?
ANSWERS
For solutions to all the problems on this page click here.
A3.
The average speed of the wrench as it passes the poster is 1.35 / 0.210 = 6.43 m/s
Since acceleration is constant, the instantaneous speed of the wrench after falling 0.105s past the top of the poster is 6.43 m/s.
Let the wrench have a speed, v_{t}, at the top of the poster, an acceleration, a = 9.81 m/s/s, and a speed, v_{m} = 6.43 m/s, at the time, t = 0.105s below the top of the poster.
Applying a = (v_{f}  v_{i})/t to the above scenario and substituting the values we obtain
a = (v_{m}  v_{t})/t
9.81 = (6.43  v_{t})/(0.105)
v_{t} = 5.40 m/s
When the wrench was initially dropped, it had a speed, v_{i} = 0 and an acceleration, a = 9.81 m/s/s. As it reached the top of the poster a distance, d, below, it had a speed of 5.40 m/s.
Applying v_{f2} = v_{i2} + 2ad to the above scenario and substituting the values we obtain
5.40_{2} = 0 _{2} + 2(9.81)d
d = 1.49 m
The wrench was released 1.49 m above the top of the poster.
For solutions to all the problems on this page click here.
B12.
a.
time to maximum height = 5.0/2 = 2.5 s
velocity at maximum height = 0
Applying a = (v_{f}  v_{i})/t yields
9.81 = (0  v_{i})/2.5
v_{i} = 24.5 m/s
b.
Applying d = v_{i}t + (0.5)at^{2} yields:
d = (24.5)(2.5) + (0.5)(9.81)(2.5)^{2} = 30.7 m
For solutions to all the problems on this page click here.
