Projectile Motion Worksheet with Problems on Horizontal Launch and Trajectory Analysis

I'll solve each problem step by step, checking my calculations carefully.

## Problem 2: Stone thrown horizontally from cliff

Given:
- Initial horizontal velocity: v₀ₓ = 8.0 m/s
- Height of cliff: h = 80 m
- Initial vertical velocity: v₀ᵧ = 0 (thrown horizontally)

Step 1: Find the time to fall 80 m vertically

Using the equation for vertical motion:
h = v₀ᵧt + ½gt²

Since v₀ᵧ = 0:
80 = 0 + ½(9.8)t²
80 = 4.9t²
t² = 80/4.9
t² = 16.33
t = √16.33
t ≈ 4.04 s

Step 2: Find horizontal distance

x = v₀ₓ × t
x = 8.0 × 4.04
x = 32.32 m

Let me verify this calculation:
t = √(2×80/9.8) = √(160/9.8) = √16.327 = 4.04 s ✓
x = 8.0 × 4.04 = 32.3 m ✓

## Problem 3: Toy car moving off table

Given:
- Height of table: h = 1.25 m
- Horizontal distance: x = 0.40 m
- Initial vertical velocity: v₀ᵧ = 0

Find: Time in air

Using vertical motion equation:
h = v₀ᵧt + ½gt²

Since v₀ᵧ = 0:
1.25 = 0 + ½(9.8)t²
1.25 = 4.9t²
t² = 1.25/4.9
t² = 0.2551
t = 0.2551
t ≈ 0.505 s

Let me verify:
t = √(2×1.25/9.8) = √(2.5/9.8) = √0.2551 = 0.505 s ✓

Rounding to 2 significant figures: t = 0.51 s

## Problem 4: Pistol fired horizontally at target

Given:
- Distance to target: x = 120 m
- Bullet velocity: v₀ₓ = 200 m/s
- Target and pistol at same height initially
- Initial vertical velocity: v₀ᵧ = 0

Part a: Time to reach target

x = v₀ × t
120 = 200 × t
t = 120/200
t = 0.60 s

Part b: How far below target does bullet hit?

During this time, the bullet falls due to gravity:
y = v₀ᵧt + ½gt²

Since v₀ = 0:
y = 0 + ½(9.8)(0.60)²
y = 4.9 × 0.36
y = 1.764 m

Let me verify:
t = 120/200 = 0.6 s ✓
y = ½ × 9.8 × (0.6)² = 4.9 × 0.36 = 1.764 m ✓

Rounding to 2 significant figures: y = 1.8 m

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Final Answer:

Problem 2: The stone will strike the ground approximately 32 m from the base of the cliff.

Problem 3: Approximately 0.51 seconds passed between when the car left the table and when it hit the floor.

Problem 4:
- Time to reach target: 0.60 seconds
- Distance below target: 1.8 m