LINEAR MOTION: FREE FALL
A detailed overview with definitions, formulas, examples, and applications
Prepared By: Ma. Johanna B. Testa, LPT
DEFINITION OF FREE FALL
Free fall refers to the motion of an object under the influence of gravitational force only, with no air resistance. All objects in free fall near Earth’s surface accelerate downward at approximately 9.8 m/s².
Source: Khan Academy, Encyclopedia Britannica
KEY FORMULAS IN FREE FALL
1. d = Vit + ½at
2. d = ½ (Vi + Vf) t
3. vf² = Vi2 + 2d
4. Vf = Vi + at
Where a = 9.8 m/s² (acceleration due to gravity)
and d= Vt (where the object is moving at a constant acceleration)
EXAMPLE PROBLEM
Problem 1: A ball is dropped from rest from the edge of a cliff. Assume the acceleration due to gravity is 9.8 m/s² and air resistance is negligible.
a. What is the speed of the ball 5 seconds later?
b. What is the velocity of the ball at this time?
c. How far does it travel during this time?
d. What is the displacement of the ball? can you rewrite the answer again into negative acceleration due to gravity
EXAMPLE PROBLEM: SOLUTION
Given:
Initial velocity 𝑉𝑖 = 0m/s
Time 𝑡 = 5s
Acceleration due to gravity 𝑎 = -9.8 m/s2
Motion is vertically downward
(a) What is the speed of the ball 5 seconds later?
Use the equation:
𝑉𝑓 = 𝑉𝑖+𝑎𝑡
𝑉𝑓 = 0 + (−9.8)(5) = −49m/s
Speed is the magnitude of velocity:
Answer: 49m/s
(b) What is the velocity of the ball at this time?
Answer: −49m/s
(The ball is moving downward, hence the negative sign.)
EXAMPLE PROBLEM: SOLUTION
Given:
Initial velocity 𝑉𝑖 = 0m/s
Time 𝑡 = 5s
Acceleration due to gravity 𝑎 = -9.8 m/s2
Motion is vertically downward
(c) How far does it travel during this time?
Use:
𝑑 = 𝑉𝑖𝑡 + ½𝑎𝑡2
𝑑 = 0 + ½(−9.8)(52) = ½(−9.8)(25) =−122.5m
Distance traveled (magnitude): 122.5m
(d) What is the displacement of the ball?
Answer: −122.5m
(The ball moved 122.5 meters downward, hence negative.)
EXAMPLE PROBLEM
Problem 2: (Thrown Ball from a cliff) A ball is thrown downward with an initial speed of 15 m/s from the top of a cliff.
(a) What is the speed and velocity of the ball 8 seconds later?
(b) How far does it travel during this time, and what is its displacement?
EXAMPLE PROBLEM: SOLUTION
Given:
Initial velocity: 𝑉𝑖 = −15m/s (downward = negative)
Time: 𝑡=8s
Acceleration: 𝑎 = −9.8m/s2
(a) Final speed and velocity after 8 seconds
Use the formula:
𝑉𝑓 = 𝑉𝑖+ 𝑎𝑡
𝑉𝑓 = −15 + (−9.8)(8) = −15−78.4 = −93.4m/s
Speed = ∣𝑉𝑓∣ =93.4m/s
Velocity = −93.4m/s (negative → downward)
EXAMPLE PROBLEM: SOLUTION
(b) Distance traveled and displacement
Use the formula:
𝑑 = 𝑉𝑖𝑡 + ½𝑎𝑡2
𝑑 = (−15)(8)+ ½(−9.8)(8)2
𝑑 = −120 + 0.5(−9.8)(64) = −120 − 313.6 = −433.6m
Distance traveled = ∣𝑑∣ = 433.6m
Displacement = −433.6m(downward)
EXAMPLE PROBLEM & SOLUTION
Problem 3. A stone is dropped from the top of a building and hits the ground 5 seconds later. How tall is the building?
Given:
Initial velocity (Vi) = 0 m/s (since the stone is "dropped")
Time (t) = 5 s
Acceleration due to gravity (a) = -9.8 m/s² (the negative sign indicates downward direction)
Unknown:
Distance (d) or height of the building
Formula:
We can use the following equation:
d = Vit + ½at2
Solution:
Substitute the given values into the formula:
d = (0m/s)(5s) + ½ (−9.8m/s2)(5s)2
d = 0 + ½(−9.8)(25)
d = −4.9 × 25
d= −122.5m
h = 122.5m (the height of the building
EXAMPLE PROBLEM & SOLUTION
Problem 4. A stone is thrown downward from the top of a cliff at 24 m/s and hits the ground 7 seconds later. How tall is the cliff?
Given:
Initial velocity (Vi) = -24 m/s (thrown downward)
Time (t) = 7 s
Acceleration due to gravity (a) = -9.8 m/s² (acting downward)
Unknown:
Distance (d) or height of the cliff
Formula:
We can use the following kinematic equation:
d = Vit + ½at2
Solution:
Substitute the given values into the formula:
d = (−24m/s)(7s) + ½(−9.8m/s2)(7s)2
d = −168 + ½(−9.8)(49)
d = −168 + (−4.9×49)
d = −168 − 240.1
d=−408.1m
The displacement calculated is -408.1 m. The negative sign indicates that the displacement is in the downward direction from the initial position. Since the question asks for the "height" of the cliff, which is a scalar quantity representing distance, we take the absolute value.
Therefore, the height of the cliff is 408.1 meters.
EXAMPLE PROBLEM & SOLUTION
Problem 5. A rock is released from rest on a 700m building. (a) How long does it take to hit the ground? (b) What is the speed and velocity of the ball just before it hits the ground?
(a) How long does it take to hit the ground?
Given:
Initial velocity (Vi) = 0 m/s (released from rest)
Distance (d) = -700 m (downward displacement, so negative)
Acceleration due to gravity (a) = -9.8 m/s² (downward)
Unknown:
Time (t)
Formula:
We use the equation:
d = Vit + ½at2
Solution:
Substitute the given values into the formula:
Therefore, it takes approximately 11.95 seconds for the rock to hit the ground.
(b) What is the speed and velocity of the ball just before it hits the ground?
Given:
Initial velocity (Vi) = 0 m/s
Acceleration (a) = -9.8 m/s²
Time (t) = 11.95 s (from part a)
Unknown:
Final velocity (Vf)
Speed
Formula:
We can use the equation:
Vf = Vi + at
Solution for Velocity:
Substitute the given values into the formula:
Vf = 0 + (−9.8m/s2)(11.95s)
Vf = −117.11m/s
Therefore, the velocity of the ball just before it hits the ground is -117.11 m/s. The negative sign indicates the downward direction.
Solution for Speed:
Speed is the magnitude of velocity.
Speed = ∣Vf∣
Speed = ∣−117.11m/s∣
Speed = 117.11m/s
Therefore, the speed of the ball just before it hits the ground is 117.11 m/s.
REAL-LIFE APPLICATIONS OF FREE FALL
- Skydiving: Understanding how gravity affects falling objects.
- Engineering: Calculating fall times for dropped tools.
- Sports Science: Studying motion in jumping events.
- Space Exploration: Analyzing gravity-based movements of landers on planetary bodies.
Source: NASA, McGraw-Hill Physics
SUMMARY
- Free fall is motion under gravity alone.
- Gravity near Earth's surface: ~9.8 m/s².
- Important equations: d = Vit + ½at, d = ½ (Vi + Vf) t, vf² = Vi2 + 2d, Vf = Vi + at
- Widely applied in science, engineering, and sports.