# Free fall motion

### Article about the Free fall motion

In everyday life, we often see objects that experience free-fall motion, for example, the motion of fruit falling from a tree, the motion of objects that fall or are dropped from a certain height. Why do objects experience free-fall motion? If observed at a glance, the object experiencing free fall as if it has a fixed speed, or in other words the object does not accelerate. The fact that happens, every object that falls freely experiences a constant acceleration. This reason causes free-fall motion, including the example of nonuniform linear motion. How to prove that objects experiencing free-fall experience constant acceleration or its speed increase?

Plug two nails in the ground and drop a stone from a different height on each nail. You will see that the nails that are subjected to stones at a higher level are stuck deeper than the other nails. The higher the stone position from the ground surface, the greater the speed of the stone when it strikes the ground so that it presses the nail deeper.

In the past, the motion of objects falling to the ground was a very intriguing subject in natural philosophy. Aristotle, a philosopher, once said that an object that has a bigger mass falls faster than lighter objects. Aristotle’s opinion influenced the views of people who lived before Galileo’s time, who considered that objects with a larger mass fall faster than lighter objects and that the speed of falling objects was proportional to the mass of the object. Maybe before learning this subject, you also think so. For example, we drop a piece of paper and a rock from the same height. The results we observed indicate that the stone first touched the ground or floor surface than paper. Now, let’s drop two stones from the same height, where one stone is bigger than the other. The two stones touched the surface of the ground at the same time, compared to the stone and paper we dropped earlier. We can also experiment with dropping stones and paper in the form of lumps.

What influences the movement of falling rocks or paper? Air friction force! Air resistance or friction greatly affects free-fall motion. Galileo postulates that all objects will fall with the same acceleration if there is no air or other obstacles. Galileo asserted that all objects, heavy or light, fall with the same acceleration, at least if there is no air. Galileo believes that air acts as an obstacle to extremely light objects that have a large surface. But in many situations, this air resistance can be ignored. In a room where the air has been sucked (vacuum), light objects such as a piece of paper held horizontally will fall with the same acceleration as other objects. Galileo’s contribution to our understanding of the motion of falling objects can be summarized as follows:

At a particular location on earth and lacking air resistance, all objects fall with the same constant acceleration. We call this acceleration caused by gravity and give it a symbol g. The magnitude of g is approximately 9.8 m/s2. In English System units, the magnitude of g is approximately 32 ft/s2. The direction of the acceleration of gravity toward the center of the earth.

### Definition of the free fall motion

An object is said to experience free fall if the object moves perpendicular to the center of the earth and during its motion, the object experiences the constant acceleration of gravity. If the free-fall near the surface of the earth, then the object experiences a constant acceleration of gravity of 9.8 m/s2

and the direction of acceleration of gravity towards the center of the earth (perpendicular to the surface of the earth). To simplify the calculations, g is 10 m/s2.

There are three different situations:

1. Objects move vertically down without initial speed (no vo). For example, the fruit falls from the tree after being released from the stem. The direction of movement is always downward and the object experiences acceleration so that g is always positive. In certain physics books called free-fall motion.

2. The object moves vertically down with the initial speed (there is vo). For example, a stone thrown vertically downward. The direction of movement is always downward and the object experiences acceleration so that g is always positive. In certain physics books, it is called vertical downward motion.

3. The object move vertically upward at an initial speed, after reaching maximum height, the object moves back downward. Suppose you throw vertical marbles up and catch them again when the marbles move down. When moving upward, objects experience a deceleration (negative g), when moving vertically down, the object accelerates (g positive). In certain physics books, it is called vertical upward motion. It should be noted that if an object experiences one of the three conditions above, the object is said to do the free-fall motion.

### The equation of the free fall motion

Free-fall motion is an example of nonuniform linear motion, therefore the equation of the free-fall motion is basically the same as the equation of the nonuniform linear motion and adjusted to the conditions in the free-fall motion.

h = height (meter), vo = initial speed (meter/second), vt = final speed (meter/second), t = time (second), g = acceleration of gravity (meter/second) = 9.8 m/s2 or 10 m/s2.

The acceleration of gravity is constant at 10 m/s2 (g positive, the object moves downward) means that the speed of the object increases by 10 m/s every 1 second. 2 seconds later, the object speed increases by 20 m/s. 3 seconds later, the object speed increases by 30 m/s. Constant deceleration of gravity is 10 m/s2 (g negative, the object moves upward) means that the speed of the object decreases by 10 m/s every 1 second. 2 seconds later, the speed of the object decreases by 20 m/s. 3 seconds later, the speed of the object decreases by 30 m/s. Constant acceleration or constant deceleration only occurs near the surface of the earth.

Sample problem 1:

Mango is released and falls to the ground. If the initial position is 10 meters from the ground surface and the mass of mango is 5 grams, determine:

(a) the speed of the mango when it arrives at the ground

(b) the time interval for the mango to reach the ground.

g = 9.8 m/s2

Solution:

Known: h = 10 m, g = 9.8 m/s2

a) The speed of manga when it arrives at the ground

Mass is calculated in the equation of the free-fall motion

b) Time interval in air

Sample problem 2:

An object is dropped from a certain height. Determine:

(a) the magnitude of the object acceleration

(b) the distance traveled by the object for the first 2 seconds

(c) the speed of the object after falling 50 meters

(d) how much time is needed for the object to reach a speed of 20 m/s,

(e) how much time is needed for the object to fall as far as 100 meters

Solution:

Known: g = 9.8 m/s2

a) The magnitude of the acceleration of the object

Acceleration of the object = acceleration of gravity = g = 9.8 m/s2

b) The distance traveled by the object for the first 2 seconds

Known: g = 9.8 m/s2 , t = 2 s

Wanted: h

h = 1⁄2 g t2 = 1⁄2 (9.8)(2)2 = (4.9)(4) = 19.6 meters

c) The speed of the object after falling as far as 50 meters

Known: h = 50 m, g = 9,8 m/s2

Wanted: vt

d) How much time is needed for objects to reach a speed of 20 m/s

Known: vt = 20 m/s, g = 9,8 m/s2

Wanted: t

e) The time interval needed for objects to fall as far as 100 meters

Wanted: h = 100 m, g = 9,8 m/s2

Solution: t

Sample problem 3 :

A stone is thrown into a well with an initial speed of 5 m/s. If the stone falls in water after 4 seconds, determine:

(a) the speed of the stone when it comes into water

(b) the depth of the well

Solution:

a) Speed of stone when it comes into water

Known:

vo = 5 m/s, t = 4 s, g = 9.8 m/s2

Wanted: vt

vt = vo + g t

vt = 5 m/s + (9.8 m/s2)(4 s) = 5 m/s + 39.2 m/s

vt = 44.2 m/s

b) Well depth

Known:

vo = 5 m/s, t = 4 s, g = 9.8 m/s2

Wanted: h

h = vo t + ½ g t2

h = (5)(4) + ½ (9.8)(4)2

h = 20 + (4.9)(16)

h = 20 + 78.4

h = 98.4 meters

Sample problem 4:

From the top of the building as high as 50 meters, a package is thrown vertically downward at a speed of 10 m/s. Determine:

(a) The time in the air

(b) The speed of the package when strikes the ground

Solution:

a) The time in air

Known:

h = 50 m, g = 9.8 m/s2, vo = 10 m/s

Wanted: t

h = vo t + ½ g t2

50 = 10 t + ½ (9.8) t2

50 = 10 t + 4.9 t2

4.9 t2 + 10 t – 50 = 0

Time in air = 2.3 seconds

b) The speed of the package when strikes the ground

Known:

h = 50 m, g = 9.8 m/s2, vo = 10 m/s

Wanted: vt

Sample problem 5:

A ball is thrown vertically upward at an initial speed of 20 m/s. Determine the maximum height reached by the ball.

Solution:

The quantities of the vector whose direction is upward are positive, the quantities of the vector whose direction is downward are negative. The initial position of the ball is chosen as the reference point.

Known:

vo = 20 m/s (direction of initial speed is upward, the ball is thrown upward, so vo is positive)

vt = 0 m/s (speed of ball at the maximum height is 0 m/s)

g = – 9.8 m/s2 (direction of the acceleration of gravity is downward, so g is negative)

Wanted: h

Sample problem 6:

A marble is thrown vertically upward from a top of a building 100 meters above the ground at an initial speed of 20 m/s. Determine:

(a) Time in air

(b) the speed of marble when strikes the ground

Solution:

Position where the marble is thrown as the reference point; the top of the building is a reference point. The magnitude of the vector whose direction is upward is positive, the magnitude of the vector whose direction is downward is negative.

a) Time in air

Known:

h = – 100 m (h is negative because the ground surface is below the initial position or reference point)

vo = 20 m/s (direction of the initial speed is upward, so vo is positive)

g = – 9.8 m/s2 (direction of the acceleration of gravity is downward, so g is negative)

Wanted: t

The time needed to reach the ground since

the ball is thrown = 7 seconds.

b) The speed of marble when strikes the surface of the ground

Known:

h = -100 m (negative because the ground surface is below the reference point or initial position)

vo = 20 m/s (the direction of the initial speed is upward, so vo is positive)

g = -9.8 m/s2 (direction of the acceleration of gravity is downward, so g is negative).

Wanted: vt

1. What is free fall motion?

Free fall motion is the motion of an object under the influence of gravity alone, with no other forces acting on it, such as air resistance.

2. What is the acceleration due to gravity?

Acceleration due to gravity is the acceleration that an object experiences when falling freely near the surface of the Earth. It is typically denoted as ‘g’ and has an approximate value of 9.8 m/s².

3. What is the formula for displacement in free fall motion, given initial velocity and time?

The formula is h = ut + ½gt², where h is displacement (height), u is initial velocity, g is acceleration due to gravity, and t is time.

4. How does the velocity of an object change during free fall motion?

During free fall, an object’s velocity increases linearly over time due to the constant acceleration of gravity.

5. What is the formula for final velocity in free fall motion, given initial velocity and time?

The formula is v = u + gt, where v is final velocity, u is initial velocity, and g is acceleration due to gravity.

6. What happens to the velocity of an object at the peak of its trajectory in free fall motion?

At the peak of its trajectory, the velocity of the object becomes zero momentarily.

7. What does a negative sign represent in the context of free fall motion?

The negative sign usually denotes a direction opposite to the chosen positive direction. Depending on the context, it can indicate falling down (if up is positive) or thrown down (if up is negative).

8. What happens to the acceleration of an object in free fall when it reaches its maximum height?

The acceleration of an object at its maximum height in free fall remains equal to g (approximately -9.8 m/s² near the Earth’s surface), pointing downward.

9. How does air resistance affect free fall motion?

In reality, air resistance can significantly slow an object’s descent, making the motion no longer a free fall. However, in many physics problems, air resistance is ignored for simplicity.

10. What is the time of flight in free fall motion?

The time of flight is the total time that the object spends in the air. For an object launched and landing at the same height, the time of flight can be found by the formula t = 2u/g.

Problems and solutions about Free fall motion

1. Problem: A rock is dropped from a cliff of height 78.4 meters. How long does it take for the rock to reach the ground? Solution: We use the equation h = ½gt². Solving for time, t = √(2h/g) = √(2×78.4/9.8) = 4 s.
2. Problem: A ball is thrown upwards with an initial velocity of 19.6 m/s. How high does the ball go? Solution: Using the equation h = v₀t – ½gt² at maximum height (where final velocity is 0), the height h = v₀² / (2g) = (19.6)² / (2×9.8) = 20 m.
3. Problem: A stone is thrown upwards with an initial velocity of 10 m/s. What will be its velocity after 2 seconds? Solution: The equation v = v₀ – gt gives velocity v = 10 – 9.8×2 = -9.6 m/s.
4. Problem: A coin is dropped down a well and hits the water after 3 seconds. How deep is the well? Solution: Using h = ½gt², the depth h = ½x9.8×3² = 44.1 m.
5. Problem: A book falls off a table and hits the ground after 0.5 seconds. How high was the table? Solution: Using h = ½gt², the height h = ½x9.8x(0.5)² = 1.225 m.
6. Problem: A ball is thrown upwards with an initial velocity of 20 m/s. When will it reach its maximum height? Solution: At maximum height, v = 0. Solving t = (v – v₀) / -g, the time t = (0 – 20) / -9.8 = 2.04 s.
7. Problem: An object falls for 6 seconds. What is its final velocity? Solution: Using v = v₀ + gt with initial velocity v₀ = 0, the final velocity v = 0 + 9.8×6 = 58.8 m/s.
8. Problem: An apple falls from a tree and takes 1.5 seconds to hit the ground. What was the height of the tree? Solution: Using h = ½gt², the height h = ½x9.8x(1.5)² = 11 m.
9. Problem: A soccer ball is kicked vertically upwards with an initial velocity of 25 m/s. How long until it hits the ground? Solution: Time to maximum height t = v₀ / g = 25 / 9.8 = 2.55 s. The total time to hit the ground is twice this, so t = 2×2.55 = 5.1 s.
10. Problem: A stone is dropped off a bridge and hits the water after 4 seconds. How high is the bridge? Solution: Using h = ½gt², the height h = ½x9.8×4² = 78.4 m.
11. Problem: A rocket is launched straight up at 50 m/s. How high does it go? Solution: Using h = v₀² / (2g), the height h = (50)² / (2×9.8) = 127.55 m.
12. Problem: A ball is thrown downwards with an initial speed of 10 m/s. What is its velocity after 2 seconds? Solution: Using v = v₀ + gt, the velocity v = 10 + 9.8×2 = 29.6 m/s.
13. Problem: A ball is thrown vertically upwards and returns to the ground in 6 seconds. What was its initial velocity? Solution: Using v₀ = gt / 2 (since total time is twice the time to reach maximum height), the initial velocity v₀ = 9.8×6 / 2 = 29.4 m/s.
14. Problem: An object falls for 10 seconds. How far does it fall? Solution: Using h = ½gt², the distance h = ½x9.8×10² = 490 m.
15. Problem: A marble is dropped from a tower and takes 5 seconds to hit the ground. How high is the tower? Solution: Using h = ½gt², the height h = ½x9.8×5² = 122.5 m.
16. Problem: A baseball is thrown vertically upwards with an initial velocity of 15 m/s. When will it reach its maximum height? Solution: At maximum height, v = 0. Solving t = (v – v₀) / -g, the time t = (0 – 15) / -9.8 = 1.53 s.
17. Problem: An object is dropped and falls for 7 seconds. What is its final velocity? Solution: Using v = v₀ + gt with initial velocity v₀ = 0, the final velocity v = 0 + 9.8×7 = 68.6 m/s.
18. Problem: A rock is thrown upwards with an initial velocity of 30 m/s. What will be its velocity after 3 seconds? Solution: Using v = v₀ – gt, the velocity v = 30 – 9.8×3 = 0.6 m/s.
19. Problem: A stone is dropped from a cliff and hits the ground after 8 seconds. How high is the cliff? Solution: Using h = ½gt², the height h = ½x9.8×8² = 313.6 m.
20. Problem: An arrow is shot straight up at 60 m/s. How long until it hits the ground? Solution: Time to maximum height t = v₀ / g = 60 / 9.8 = 6.12 s. The total time to hit the ground is twice this, so t = 2×6.12 = 12.24 s.