A projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high (as shown in the figure(Figure 1)).If the projectile lands on top of the cliff 6.8 s after it is fired, find the initial velocity of the projectile ( (a)magnitude and (b)direction ). Neglect air resistance.

Answer

Projectile Motion: Calculating Initial Velocity to Reach a Cliff – Learnlyfly

Projectile Motion: Calculating Initial Velocity to Reach a Cliff

In projectile motion problems, finding the initial velocity involves understanding both horizontal and vertical components of motion. When a projectile is launched from ground level to land on top of a cliff, we must apply the principles of 2D kinematics. Here, we determine both the magnitude and direction of the launch velocity.

📊 Given Data

Horizontal distance to the cliff, x = 195 m
Vertical height of the cliff, y = 135 m
Time of flight, t = 6.8 s
Acceleration due to gravity, g = 9.8 m/s²
Air resistance: Neglected

📐 Step 1: Break Motion into Components

The motion is two-dimensional, so we split the velocity into:

v_0x: Horizontal component
v_0y: Vertical component

➡️ Horizontal Motion

Horizontal velocity remains constant (no acceleration):

x = v_0x × t ⇒ v_0x = x / t = 195 / 6.8 ≈ 28.68 m/s

⬆️ Vertical Motion

Use the kinematic equation:

y = v_0y × t – (1/2)gt²

Substitute the known values:

135 = v_0y × 6.8 – 0.5 × 9.8 × (6.8)²

Simplify:

135 = v_0y × 6.8 – 0.5 × 9.8 × 46.24 ≈ v_0y × 6.8 – 226.97

v_0y × 6.8 = 135 + 226.97 = 361.97

v_0y ≈ 361.97 / 6.8 ≈ 53.23 m/s

🧮 Step 2: Find the Magnitude of Initial Velocity

Combine horizontal and vertical components using the Pythagorean theorem:

v₀ = √(v_0x² + v_0y²) = √(28.68² + 53.23²) = √(823.87 + 2833.38) = √(3657.25) ≈ 60.48 m/s

✅ Magnitude of initial velocity: ≈ 60.5 m/s

🧭 Step 3: Find the Direction of Initial Velocity

Use trigonometry to find the launch angle θ:

θ = tan⁻¹(v_0y / v_0x) = tan⁻¹(53.23 / 28.68) ≈ tan⁻¹(1.855) ≈ 61.4°

✅ Direction of launch: ≈ 61.4° above the horizontal

📘 Summary of Final Results

Initial horizontal velocity (v_0x): 28.68 m/s
Initial vertical velocity (v_0y): 53.23 m/s
Magnitude of initial velocity: 60.5 m/s
Direction (angle of launch): 61.4° above the horizontal

📈 Physics Concepts Applied

✔️ 2D Kinematic Equations: Break motion into x and y components.
✔️ Uniform Horizontal Motion: No acceleration in x-direction.
✔️ Vertical Motion with Gravity: Acceleration due to gravity affects y-direction.
✔️ Pythagoras & Trig: Used to find vector magnitude and angle.

🎓 Real-World Applications

🛰️ Satellite launches and ballistic missiles
🏀 Sports like basketball, soccer, and javelin throw
🎯 Calculating the path of projectiles in video games

💬 Conclusion

By breaking the motion into components and applying basic kinematic formulas, we were able to calculate both the initial velocity’s magnitude and direction for a projectile landing on a cliff. This method is widely applicable in both real-life physics problems and standardized exams. Practice similar problems to master the vector-based approach to projectile motion!

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