Motion in a Plane - Projectile Motion Solved Examples (Class 11 Physics)
Analyze projectile motion by resolving it into horizontal and vertical components. These problems cover range, maximum height, time of flight, and trajecto
TL;DR: Analyze projectile motion by resolving it into horizontal and vertical components. These problems cover range, maximum height, time of flight, and tra…
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Analyze projectile motion by resolving it into horizontal and vertical components. These problems cover range, maximum height, time of flight, and trajecto
Motion in a Plane - Projectile Motion — Solved Numerical Examples (Step by Step)
Example 1: A projectile is thrown horizontally from a cliff with velocity 20 m/s. The cliff is 80 m high. Find the time of flight and horizontal range. (g = 10 m/s²)
Solution: Vertical motion (initial vertical velocity = 0):
h = (1/2)gt²
80 = (1/2) × 10 × t²
80 = 5t²
t² = 16
t = 4 s (time of flight)
Horizontal range = horizontal velocity × time = 20 × 4 = 80 m
Example 2: A ball is projected at 45° with initial velocity 40 m/s. Find the maximum height and range. (g = 10 m/s²)
Solution: Vertical component u_y = 40 sin(45°) = 40 × (1/√2) = 40/√2 = 20√2 m/s
Horizontal component u_x = 40 cos(45°) = 40 × (1/√2) = 20√2 m/s
Maximum height H = u_y²/(2g) = (20√2)² / (2 × 10) = 800/20 = 40 m
Range R = (u² sin(2θ))/g = (40² × sin(90°))/10 = 1600 × 1 / 10 = 160 m
Example 3: A projectile is launched at an angle of 30° with initial velocity 60 m/s. Find the time of flight. (g = 10 m/s²)
Solution: Vertical component u_y = 60 sin(30°) = 60 × 0.5 = 30 m/s
Time of flight T = (2 × u_y) / g = (2 × 30) / 10 = 6 s
Example 4: A projectile reaches a maximum height of 20 m. If it is projected at 60°, find the initial velocity. (g = 10 m/s²)
Solution: At maximum height, vertical velocity = 0
u_y = u sin(60°) = u × (√3/2)
Using v² = u² - 2gs (at max height, v = 0)
0 = u_y² - 2 × 10 × 20
u_y² = 400
u_y = 20 m/s
Initial velocity u = u_y / sin(60°) = 20 / (√3/2) = 40/√3 = 40√3/3 ≈ 23.1 m/s
Example 5: A stone is thrown from ground level at 53° with velocity 50 m/s. What is the height of the stone when it is at a horizontal distance of 100 m? (g = 10 m/s²)
Solution: u_x = 50 cos(53°) = 50 × 0.6 = 30 m/s
u_y = 50 sin(53°) = 50 × 0.8 = 40 m/s
Time to reach horizontal distance 100 m: t = 100 / 30 = 10/3 s
Height y = u_y × t - (1/2)g × t²
y = 40 × (10/3) - (1/2) × 10 × (10/3)²
y = 400/3 - 5 × 100/9 = 400/3 - 500/9 = 1200/9 - 500/9 = 700/9 ≈ 77.8 m
Example 6: For a projectile, the time of flight is 4 seconds and range is 200 m. Find the angle of projection. (g = 10 m/s²)
Solution: Time of flight T = (2u sin(θ))/g
4 = (2u sin(θ))/10
40 = 2u sin(θ)
u sin(θ) = 20 ... (1)
Range R = (u² sin(2θ))/g = (u² × 2 sin(θ) cos(θ))/g
200 = (u² × 2 sin(θ) cos(θ))/10
2000 = 2u² sin(θ) cos(θ)
1000 = u × (u sin(θ)) × cos(θ)
1000 = u × 20 × cos(θ) (using equation 1)
50 = u cos(θ) ... (2)
Dividing (1) by (2): tan(θ) = 20/50 = 2/5 = 0.4
θ = tan⁻¹(0.4) ≈ 21.8°
Tips
- Resolve initial velocity into horizontal (u cos θ) and vertical (u sin θ) components; horizontal component remains constant.
- For maximum range, the angle of projection is 45°. For equal ranges, angles are θ and (90° - θ).
- At maximum height, vertical velocity is zero, but horizontal velocity remains constant.
Frequently Asked Questions
Why is the angle of 45° best for maximum range?
Range R = (u² sin(2θ))/g is maximum when sin(2θ) = 1, which occurs at 2θ = 90° or θ = 45°.
Does air resistance affect projectile motion in reality?
Yes, air resistance significantly affects real projectiles by reducing range and height. Our calculations assume an ideal case without air resistance.
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