Topic 5: Projectile Motion

Course: Prep4Uni Physics 1

Chapter 1: Quantities and Measurement
Chapter 2: Forces and Moments
Chapter 3: Motion and Forces
Chapter 4: Energy and Fields
Chapter 5: Projectile Motion

🚁Overview

This topic focuses on the motion of objects under the influence of gravity, especially when launched with an initial horizontal or angled velocity. It also introduces the effects of air resistance and the concept of terminal velocity, helping you develop both qualitative and quantitative understanding of real-world projectile motion.

📖Contents

Free Fall and Gravitational Acceleration
Gravitational Potential Energy
Analyzing Projectile Motion
Air Resistance and Terminal Velocity
Qualitative Description of Motion with Air Resistance

🎯Learning Outcomes

By the end of this section, students should be able to:

Understand weight as a force in a gravitational field
Analyze motion with perpendicular velocity and acceleration components
Derive and use the gravitational potential energy equation: ΔEp = mgΔh
Describe the qualitative effects of air resistance on motion
Understand the concept of terminal velocity

🔢1. Free Fall and Gravitational Acceleration

Key Idea:
In the absence of air resistance, all objects fall with the same acceleration due to gravity (approximately 9.8 m/s² on Earth), regardless of mass.

Weight (W):
W = mg
Where m = mass (kg), g = gravitational field strength (≈ 9.8 N/kg)

Example:
A 3 kg object in free fall experiences a weight of:
W = 3 × 9.8 = 29.4 N

📐Learning-Check 1:
Q: What is the weight of a 4 kg object on Earth?
A: W = mg = 4 × 9.8 = 39.2 N

📐Learning-Check 2:

Q: True or False: A heavier object falls faster than a lighter one in free fall.
A: False — In a vacuum, all objects fall at the same rate.

🔢2. Gravitational Potential Energy

Gravitational Potential Energy (GPE):
ΔE_p = mgΔh
Where Δh = change in height

Energy conversion during free fall:

GPE decreases
Kinetic energy increases
Total mechanical energy remains constant (ignoring air resistance)

Example:
A 1.5 kg rock falls 10 m.
ΔE_p = 1.5 × 9.8 × 10 = 147 J (converted to kinetic energy)

📐Learning-Check 1:
Q: A 0.8 kg object is lifted 2.5 m. What is its gain in potential energy?
A: ΔE_p = mgh = 0.8 × 9.8 × 2.5 = 19.6 J

📐Learning-Check 2:
Q: What energy change occurs as an object falls freely?
A: Gravitational potential energy decreases; kinetic energy increases

🔢3. Analyzing Projectile Motion

Key Concepts:

Horizontal and vertical motions are independent
Horizontal motion: constant velocity (no horizontal force)
Vertical motion: accelerated motion (gravity causes downward acceleration)

Equations used (for vertical motion):

v = u + at
s = ut + ½at²
v² = u² + 2as

Derivation of the vertical motion equation v = u + at

This equation relates final velocity (v) to initial velocity (u), acceleration (a), and time (t) for motion under constant acceleration — such as vertical motion under gravity (where a = g)

$Assumptions:$

- Motion is linear (vertical in this case)
- Acceleration is constant
- No air resistance

Definition of acceleration

Acceleration is the rate of change of velocity with respect to time:

$a = \frac{\Delta v}{\Delta t} = \frac{v – u}{t}$

Where:

- - v = final velocity
  - u = initial velocity
  - t = time taken
  - a = constant acceleration

Rearranging the equation

Multiply both sides by t: v – u = at

$t$

Solve for v

Add $u$ to both sides:

$v = u + at$

📌 Interpretation in Vertical Motion

If an object is thrown upward, $a = - g$ (acceleration acts downward)
If an object is falling, $a = + g$ (acceleration in the same direction as motion)

Example:
A ball is dropped from rest. Find its velocity after 3 seconds.
$u = 0$ , $a = 9.8 m/s^{2}$ , t=3

$v = u + at = 0 + 9.8 \cdot 3 = \boxed{29.4\, \text{m/s downward}}$

Derivation of the vertical motion equation s = ut + ½at²

Assumptions:

- Motion is in a straight line (vertical in this case)
- Acceleration is constant (e.g., gravity)
- Displacement starts from zero at t = 0

Start with the definition of average velocity:

Average velocity = (initial velocity + final velocity) / 2
So,

v_avg = (u + v) / 2

Displacement is average velocity × time:

s = v_avg × t = [(u + v)/2] × t

Substitute the expression for final velocity from the first equation of motion:

We already derived that:
v = u + at

Substitute into the equation for s:

s = [(u + (u + at)) / 2] × t
s = [(2u + at) / 2] × t

Expand the expression:

s = (2u + at) × t / 2
s = (2ut + at²) / 2

Final result:

s = ut + ½at²

Derivation of of the vertical motion equation

v² = u² + 2as

Displacement s is given by the average velocity multiplied by time:
s = v_average × t

The average velocity is given by:
v_average = (u + v) ÷ 2

So,

$s = \frac{(u + v)}{2} \cdot t$

From the first equation of motion:
v = u + at

Rearranging,
t = (v − u) ÷ a

Substitute this expression for t into the displacement formula:

$s = \frac{(u + v)}{2} \cdot \frac{(v – u)}{a}$

Multiply the two brackets using the identity
(a + b)(a − b) = a² − b²:

$s = \frac{v^2 – u^2}{2a}$

Multiply both sides by 2a:

$2as = v^2 – u^2$

Rearranging gives:

$v^2 = u^2 + 2as$

Example:
A ball is thrown horizontally at 10 m/s from a 20 m high platform.
Time to hit the ground:
s = ½gt² → t² = 2s/g = 40/9.8 → t ≈ 2.02 s
Horizontal range = vt = 10 × 2.02 ≈ 20.2 m

Path of Ball horizontal speed 10 and height 20 — Path of Ball horizontal speed 10 m/s and height 20 m

📐Learning-Check 1:
Q: A ball is projected horizontally at 12 m/s and hits the ground after 3 s. How far did it travel horizontally?
A: Horizontal distance = vt = 12 × 3 = 36 m

📐Learning-Check 2:
Q: In projectile motion, what causes vertical acceleration?
A: Gravity

🔢4. Air Resistance and Terminal Velocity

Air Resistance:

A drag force that increases with speed
Opposes motion, reducing acceleration

Terminal Velocity:

Occurs when air resistance = weight
Net force = 0 → object falls at constant speed

Example:
A skydiver accelerates initially but eventually reaches terminal velocity when upward air resistance equals their weight.

📐Learning-Check 1:
Q: What happens to net force when terminal velocity is reached?
A: Net force becomes zero

📐Learning-Check 2:
Q: Why do objects with larger surface areas reach lower terminal velocities?
A: They experience greater air resistance

🔢5. Qualitative Description of Motion with Air Resistance

Without Air Resistance (Ideal Case):

Projectile follows a symmetric parabolic path

With Air Resistance:

Path becomes asymmetric
Maximum height and range are reduced
Time of flight is affected

Projectile paths with and without air resistance

📐Learning-Check 1:
Q: Describe how air resistance changes the trajectory of a projectile.
A: It shortens the range and reduces the maximum height

📐Learning-Check 2:
Q: How does a projectile’s downward motion differ with air resistance?
A: It accelerates less rapidly than in a vacuum due to upward drag

⚙️Key Concepts Recap

Concept	Formula	Notes
Weight	W = mg	Force due to gravity
GPE	ΔE_p = mgΔh	Change in vertical position
Free fall	a = g ≈ 9.8 m/s²	Uniform vertical acceleration
Projectile range	Range = v_x × t	Horizontal motion is uniform
Terminal velocity	F_air = mg	Net force = 0 → constant speed

Go back to Prep4Uni Physics 1

📝EXERCISES

Questions and Answers

🟢 Section 1: Free Fall and Gravitational Acceleration

1. What is the value of acceleration due to gravity near Earth’s surface?
Answer: Approximately 9.8 m/s² downward.

2. An object is dropped from rest. How long will it take to fall 19.6 m?
Answer: Use s = ½gt² → t² = 2s/g = 39.2 / 9.8 = 4 → t = 2 s

3. A 3 kg object is in free fall. What is its weight?
Answer: W = mg = 3 × 9.8 = 29.4 N

4. True or False: A heavier object falls faster than a lighter one in a vacuum.
Answer: False – All objects fall at the same rate without air resistance.

5. What kind of motion does an object in free fall undergo?
Answer: Uniformly accelerated motion under gravity.

🟢 Section 2: Gravitational Potential Energy

6. Write the equation for gravitational potential energy.
Answer: ΔEp = mgΔh

7. A 2 kg object is lifted from 1 m to 4 m. What is the gain in potential energy?
Answer: ΔEp = 2 × 9.8 × (4 − 1) = 58.8 J

8. If a 5 kg rock falls 6 m, how much GPE is converted to kinetic energy?
Answer: ΔEp = 5 × 9.8 × 6 = 294 J

9. What energy transformation occurs during free fall?
Answer: Gravitational potential energy is converted to kinetic energy.

10. At the highest point of a vertical throw, what is the object’s KE and GPE?
Answer: KE = 0, GPE = maximum.

🟢 Section 3: Analyzing Projectile Motion

11. What are the horizontal and vertical components of motion in a projectile?
Answer: Horizontal: constant velocity, Vertical: accelerated motion (by gravity)

12. A ball is thrown horizontally at 10 m/s from a 20 m high cliff. How far will it land from the base?
Answer:
t = √(2h/g) = √(40 / 9.8) ≈ 2.02 s
Range = 10 × 2.02 ≈ 20.2 m

13. True or False: Gravity affects only the vertical component of projectile motion.
Answer: True

14. What is the vertical acceleration of a projectile at the top of its path?
Answer: 9.8 m/s² downward (still under gravity)

15. A projectile is launched at an angle. What shape does its path take (ideal case)?
Answer: A symmetric parabola

🟢 Section 4: Air Resistance and Terminal Velocity

16. What is air resistance?
Answer: A frictional force opposing motion through air.

17. What is terminal velocity?
Answer: The constant speed reached when air resistance equals weight.

18. True or False: Terminal velocity occurs when net force becomes zero.
Answer: True

19. What two forces act on a falling object with air resistance?
Answer: Weight (downward) and air resistance (upward)

20. Why do skydivers initially accelerate and then fall at constant speed?
Answer: Initially weight > air resistance → accelerates; then they balance → terminal velocity.

🟢 Section 5: Qualitative Effects of Air Resistance

21. Describe the effect of air resistance on the range of a projectile.
Answer: Reduces the range — projectile travels a shorter distance.

22. How does air resistance affect time of flight?
Answer: Usually increases slightly for vertical drops, decreases for angled launches.

23. What happens to the trajectory shape due to air resistance?
Answer: It becomes asymmetric — steeper descent than ascent.

24. Why does a feather fall slower than a stone in air?
Answer: The feather experiences greater air resistance relative to its weight.

25. How does surface area affect terminal velocity?
Answer: Larger surface area → more air resistance → lower terminal velocity

Problems and Solutions

1. A stone is dropped from rest from a cliff. How fast is it moving after 4 seconds?
Solution:
Given: u = 0, a = 9.8 m/s², t = 4 s
Using v = u + at:
v = 0 + 9.8 × 4 = 39.2 m/s

2. A ball is thrown upward at 20 m/s. How high does it rise before stopping momentarily?
Solution:
At the top, v = 0, u = 20 m/s, a = –9.8 m/s²
Using v² = u² + 2as:
0 = 400 + 2(–9.8)(s) → s = 400 / 19.6 = 20.41 m

3. How long does it take the ball in Q2 to reach the highest point?
Solution:
v = u + at → 0 = 20 – 9.8t → t = 20 / 9.8 = 2.04 s

4. A car accelerates uniformly from rest at 2 m/s². How far does it travel in 10 s?
Solution:
u = 0, a = 2 m/s², t = 10 s
Using s = ut + ½at²:
s = 0 + 0.5 × 2 × 100 = 100 m

5. A rock falls freely and travels 45 m. Find its final speed before hitting the ground.
Solution:
u = 0, s = 45 m, a = 9.8 m/s²
v² = u² + 2as = 0 + 2 × 9.8 × 45 = 882 → v = √882 ≈ 29.7 m/s

6. A runner slows down from 8 m/s to 2 m/s with a uniform deceleration of 1.5 m/s². How long does it take?
Solution:
v = u + at → 2 = 8 + (–1.5)t → t = (2 – 8)/–1.5 = 4 s

7. A projectile is launched vertically with initial speed 15 m/s. Find the time it takes to return to the ground.
Solution:
Time to rise = t = u / g = 15 / 9.8 ≈ 1.53 s
Total time = 2 × 1.53 = 3.06 s

8. What is the total displacement of the projectile in Q7 after 3.06 s?
Solution:
It returns to the ground → displacement = 0 m

9. A ball is dropped from 30 m high. How long does it take to hit the ground?
Solution:
s = ½gt² → 30 = 0.5 × 9.8 × t² → t² = 30 / 4.9 = 6.12 → t ≈ 2.47 s

10. A bullet is fired upward at 50 m/s. How long before it is 60 m above the launch point?
Solution:
Use s = ut + ½at² → 60 = 50t – 4.9t²
Rearranged: 4.9t² – 50t + 60 = 0
Solve with quadratic formula:
t ≈ 1.34 s or 9.11 s (ascending and descending)

11. A rocket goes from 0 to 60 m/s in 6 s. What is its acceleration and distance covered?
Solution:
a = (v – u)/t = (60 – 0)/6 = 10 m/s²
s = ut + ½at² = 0 + 0.5 × 10 × 36 = 180 m

12. A cyclist decelerates from 12 m/s to rest in 4 s. What is the distance covered?
Solution:
a = (0 – 12)/4 = –3 m/s²
s = ut + ½at² = 12 × 4 + 0.5 × (–3) × 16 = 48 – 24 = 24 m

13. A ball is thrown downward at 5 m/s from a height of 25 m. What is its speed just before hitting the ground?
Solution:
v² = u² + 2as = 25 + 2 × 9.8 × 25 = 25 + 490 = 515 → v ≈ 22.7 m/s

14. An object takes 3 seconds to hit the ground after falling. What height did it fall from?
Solution:
s = ut + ½gt² = 0 + 0.5 × 9.8 × 9 = 44.1 m

15. A motorbike speeds up from 15 m/s to 30 m/s in 5 seconds. Find its acceleration and distance covered.
Solution:
a = (30 – 15)/5 = 3 m/s²
s = ut + ½at² = 15 × 5 + 0.5 × 3 × 25 = 75 + 37.5 = 112.5 m

16. A falling object reaches 40 m/s after falling 50 m. What was its initial velocity?
Solution:
v² = u² + 2as → 1600 = u² + 2 × 9.8 × 50 = u² + 980 → u² = 620 → u ≈ 24.9 m/s downward

17. A diver jumps vertically upwards at 6 m/s. What maximum height does she reach?
Solution:
v² = u² + 2as → 0 = 36 – 2 × 9.8 × s → s = 36 / 19.6 ≈ 1.84 m

18. A particle is projected upward and reaches a maximum height after 2.5 s. What was its initial velocity?
Solution:
v = u + at → 0 = u – 9.8 × 2.5 → u = 24.5 m/s

19. How long does it take an object thrown upward at 10 m/s to fall back to the same level?
Solution:
Time to rise = 10 / 9.8 = 1.02 s
Total time = 2.04 s

20. A stone is thrown vertically upward and reaches a height of 45 m. What was its launch speed?
Solution:
v² = u² + 2as → 0 = u² – 2 × 9.8 × 45 → u² = 882 → u ≈ 29.7 m/s

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