Projectile Motion: PECTAA Class 11 Physics, Punjab Board

A ball leaves the bat at an angle and lands far away. Two questions confuse students: why does it curve, and how far will it go? Split the motion into a sideways part and an up-down part, and both answers fall out of equations you already know.

Chapter 2 of the PECTAA Physics 11 book is Force and Motion. Its projectile motion part builds the three results every paper asks for: maximum height, time of flight and range. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

What a projectile is. The book's examples are a football kicked by a player, a ball thrown by a cricketer and a missile fired from a launching pad. Each is launched at some angle to the horizontal. The book describes projectile motion as two-dimensional motion under constant acceleration due to gravity.

Split the launch velocity. Launch with speed vᵢ at angle θ above the horizontal. Then

• horizontal part: vᵢₓ = vᵢ cos θ
• vertical part: vᵢᵧ = vᵢ sin θ

Horizontal motion. With air resistance neglected, no force acts sideways, so the horizontal acceleration is zero. The horizontal velocity stays vᵢ cos θ for the whole flight (equation 2.18 in the book).

Vertical motion. Gravity acts downward, so the vertical acceleration is −g. Using the first equation of motion, the vertical velocity at time t is vᵧ = vᵢ sin θ − gt (equation 2.19). The size of the velocity at any instant comes from combining the two parts by Pythagoras, and its direction from tan of the angle = vᵧ / vₓ.

Maximum height. At the top the vertical velocity is zero. Using 2aS = v_f² − vᵢ² with a = −g:

h = vᵢ² sin²θ / 2g (equation 2.22)

Time of flight. The body returns to launch level, so the vertical displacement is zero. Solving gives

t = 2vᵢ sin θ / g (equation 2.23)

Range. Horizontal velocity × time of flight:

R = vᵢ cos θ × 2vᵢ sin θ / g = (vᵢ² / g) sin 2θ (equation 2.24)

Maximum range. sin 2θ is largest (equal to 1) when 2θ = 90°, so θ = 45°.

Air resistance. The book explains that air resistance lowers the upward velocity, so the height falls. It also slows the forward motion, so the range falls. Because air resistance changes as the body slows, the real path is not a perfect parabola.

Key points

• Two motions, one clock. The sideways and up-down motions are separate, but they share the same time t.
• Sideways speed is constant (no air resistance). Only the vertical velocity changes.
• Angles below and above 45°. The book's side box notes that below 45° both height and range are less; above 45° the height is more but the range is again less.

Worked examples

Example 1 (book Example 2.6). A ball is thrown at 30 m s⁻¹ at 30° above the horizontal. Find the height, time of flight and range.

1. vᵢₓ = 30 × cos 30° = 25.98 m s⁻¹ and vᵢᵧ = 30 × sin 30° = 15 m s⁻¹.
2. Time of flight: t = 2 × 15 / 9.8 ≈ 3.1 s.
3. Height: h = (30)² × (0.5)² / (2 × 9.8) = 225 / 19.6 ≈ 11.5 m.
4. Range: R = (30)² × sin 60° / 9.8 = 900 × 0.866 / 9.8 ≈ 79.53 m.

Example 2 (same speed, best angle). Using the same 30 m s⁻¹, what is the largest possible range?

1. Use θ = 45°, so sin 2θ = sin 90° = 1.
2. R = 900 / 9.8 ≈ 91.8 m. This is more than the 79.53 m at 30°.

Example 3 (time to the top). For Example 2.6, how long does the ball take to reach its highest point?

1. At the top vᵧ = 0, so 0 = 15 − 9.8t.
2. t = 15 / 9.8 ≈ 1.53 s, which is half the time of flight.

Common mistakes in the exam

1. Putting mass in the formulas. Height, time and range have no mass term. A heavier ball does not go farther in this model.
2. Letting the sideways speed change. Without air resistance it stays vᵢ cos θ throughout.
3. Using cos θ in the height formula. Height and time use the vertical part, vᵢ sin θ.
4. Saying vertical acceleration is zero at the top. Only the vertical velocity is zero there. The acceleration is still g downward.
5. Forgetting the factor 2 in time of flight. Up and down take equal time.
6. Writing 2 sin θ cos θ without simplifying. It equals sin 2θ, which gives the 45° result.

Exam technique

• Draw the launch velocity and its two components before writing formulas.
• In derivations, name the equation of motion you use for each step.
• Keep g = 9.8 m s⁻² unless the paper gives another value, and round only at the end.
• Check your board's notice for the topics in your paper.

Practice questions

1. A stone is thrown at 20 m s⁻¹ at 45°. Find its range. Answer: R = 400 × 1 / 9.8 ≈ 40.8 m.
2. For the stone in question 1, find the time of flight. Answer: t = 2 × 20 × 0.707 / 9.8 ≈ 2.89 s.
3. What is the vertical velocity of a projectile at its highest point? Answer: Zero. The horizontal velocity is still vᵢ cos θ.
4. Which angle gives the maximum range, and why? Answer: 45°, because sin 2θ = sin 90° = 1, its largest value.
5. State one effect of air resistance on a projectile. Answer: It lowers the height and reduces the range, so the path falls short of a perfect parabola.

Ready to practise this? Continue in 1st Year Physics.