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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/s². (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_fv_it 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.

a. Now we can see we need a relationship between d, a, v_i, and t
d = v_it + (0.5)at²
(7.0 m) = (0)t + (0.5)(9.8)t²
t = 1.20 s

b. We need a relationship between d, a, v_i, and v_f

v_f² = v_i² + 2ad

v_f² = 0² + 2(9.8)(7.0)

v_f = 11.7 m/s

THROW-UP 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 = +

* 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_it + (0.5)at²

(7.0 m) = (-2.00)t + (0.5)(9.8)t²

t = {1.42, -1.01}

Since t < 0 has no meaning,

t = 1.42 s

CATCH-UP PROBLEMS
In catch-up 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_fv_it 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.

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_it + (0.5)at²

8.0 = (0)t + (0.5)(9.8 m/s2)t²

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:

For the rock we need a relationship between d, a, v_i, and t

d = v_it + (0.5)at²

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

1. Aunt Minnie drops a penny into a wishing well and it falls for 3 seconds before hitting the water. What is the penny's average speed during its 3 second drop?

2. A rock is dropped from the top of a cliff and strikes the ground 6.5 seconds later. How high is the cliff in meters?

3. It takes 0.210s for a dropped object to pass a window that is 1.35 meters tall. From what height above the top of the window was the ball dropped?

4. What is the acceleration of a falling sky diver (100 kg including parachute) when the upward force of air resistance is equal to one-fourth of his weight?

B. Throw Up Problems

1. You throw a ball downward from a window at a speed of 2.0 m/s. The ball accelerates at 9.8 m/s².

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. At the beginning of a basketball game, a referee tosses the ball straight up with a speed of 4.6 m/s. A player cannot touch the ball until after it reaches its max height and begins to fall down. What is the minimum time that a player must wait before touching the ball.

3. A chinook salmon has a maximum underwater speed of 3.58 m/s, but it can jump out of water with a speed of 6.26 m/s. To move upstream past a waterfall, the salmon does not need to jump to the top of the fall, but only to a point in the fall where the water speed is less than 3.58 m/s; it can then swim up the fall for the remaining distance. Because the salmon must make forward progress in the water, lets assume that it can swim to the top if the water speed is 3.00 m/s. If the water has a speed of 1.50 m/s as it passes over a ledge, how far below the ledge will the water be moving with a speed of 3.00 m/s? If the salmon is able to jump vertically upward from the base of the fall, what is the maximum height of waterfall that the salmon can clear?

4. How high will an arrow be 7 seconds after being shot straight up at 50 m/s?

5. A ball is thrown straight upward and rises to a maximum height of 24 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?

6. A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s and the stone is 1.50 m above the ground when launched.

a. How high above the ground does the stone rise?
b. How much time elapses before the stone hits the ground?

7. A helium balloon is hovering at rest above the ground. The balloonist then cuts a sandbag free and the balloon begins to move upwards with a constant velocity of 2 m/s while the sandbag falls to the ground below. When the sandbag hits, the balloon is 50 m above the ground.
a. How far above the ground was the balloon when the sandbag was first released?
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?

10.

a. You shoot an arrow straight up at 50 m/s. When will it run out of speed?

b. So what will be the arrow's speed 7 seconds?

11. A bullet is fired straight up with a muzzle velocity of 460 m/s. How long will it take it to reach its highest point and how high will that be? (air resistance may be neglected.)

12. A ball thrown vertically upward is caught by the thrower after 5.00 s. (a) Find the initial velocity of the ball. (b) Find the maximum height it reaches.

C. Catch Up Problems

1. A boy shoots a rock with an initial velocity of 21 m/s straight up from his slingshot. He quickly reloads and shoots another rock in the same way 3.0 s later. At what time and height do the rocks meet?

2. A hot air balloon has just lifted off and is rising at the constant rate of 2.03 m/s. Suddenly, one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward. If the passenger is 2.50 m above her friend when the camera is tossed, what is the minimum initial speed of the camera if it is to just reach the passenger?

D. Other Problems

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 object as it passes the window is 1.35 / 0.210 = 6.43 m/s

Since acceleration is constant, the instantaneous speed of the object after falling 0.105s past the top of the window is 6.43 m/s.

Let the object have a speed, v_t, at the top of the window, 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 window.

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 object 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 window 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 object was released 1.49 m above the top of the window.

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_it + (0.5)at² 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.