Solved Name: Horizontal Projectile Motion Worksheet. 1. A | Chegg.com - Free Printable

Let’s solve each problem step by step.

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Problem 1:

A stone is thrown horizontally from a cliff 30m high with an initial speed of 20 m/s.

We know:
- Vertical height (h) = 30 m
- Initial horizontal velocity (v_x) = 20 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Initial vertical velocity (v_y0) = 0 (since it’s thrown horizontally)

Part a: How far from the cliff does the stone strike the ground?

First, find time in air using vertical motion:

Use formula:
> h = (1/2) * g * t²
→ 30 = 0.5 * 9.8 * t²
→ 30 = 4.9 * t²
→ t² = 30 / 4.9 ≈ 6.122
→ t ≈ √6.122 ≈ 2.474 seconds

Now, horizontal distance = v_x * t = 20 * 2.474 ≈ 49.48 meters

✔ Answer for part a: ~49.5 meters

Part b: What is the velocity (speed and direction) as it hits the ground?

Horizontal velocity remains constant: v_x = 20 m/s

Vertical velocity at impact:
> v_y = g * t = 9.8 * 2.474 ≈ 24.25 m/s downward

Speed = √(v_x² + v_y²) = √(20² + 24.25²) = √(400 + 588.06) = √988.06 ≈ 31.43 m/s

Direction: angle below horizontal θ = tan⁻¹(v_y / v_x) = tan⁻¹(24.25 / 20) = tan⁻¹(1.2125) ≈ 50.5° below horizontal

✔ Answer for part b: Speed ≈ 31.4 m/s, direction ≈ 50.5° below horizontal

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Problem 2:

Ball rolls off table 1.44m above floor, lands 2m horizontally away.

Part a: Time in air?

Use vertical motion:
> h = (1/2)gt² → 1.44 = 4.9 * t²
→ t² = 1.44 / 4.9 ≈ 0.2939
→ t ≈ √0.2939 ≈ 0.542 seconds

✔ Part a: ~0.54 s

Part b: Initial velocity?

Horizontal distance = v_x * t → 2 = v_x * 0.542
→ v_x = 2 / 0.542 ≈ 3.69 m/s

✔ Part b: ~3.69 m/s

Part c: Velocity when hitting floor?

v_x = 3.69 m/s (constant)
v_y = g*t = 9.8 * 0.542 ≈ 5.31 m/s down

Speed = √(3.69² + 5.31²) = √(13.62 + 28.20) = √41.82 ≈ 6.47 m/s

Angle = tan⁻¹(5.31 / 3.69) ≈ tan⁻¹(1.44) ≈ 55.2° below horizontal

✔ Part c: ~6.47 m/s at ~55.2° below horizontal

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Problem 3:

Cat jumps off 1.5m high table with initial velocity 10 m/s horizontally.

Time in air:
> h = 0.5*g*t² → 1.5 = 4.9*t²
→ t² = 1.5 / 4.9 ≈ 0.306
→ t ≈ √0.306 ≈ 0.553 s

Horizontal distance = v_x * t = 10 * 0.553 ≈ 5.53 meters

✔ Answer: ~5.53 meters

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Problem 4:

Bomber flying at 9 km height, speed 800 km/h. How far before target to drop bombs?

Convert units:

Height = 9000 m
Speed = 800 km/h = 800,000 m / 3600 s ≈ 222.22 m/s

Time to fall:
> h = 0.5*g*t² → 9000 = 4.9*t²
→ t² = 9000 / 4.9 ≈ 1836.73
→ t ≈ √1836.73 ≈ 42.86 seconds

Horizontal distance = v_x * t = 222.22 * 42.86 ≈ 9524 meters or 9.52 km

✔ Answer: ~9.52 km before target

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Problem 5:

Arrow shot horizontally at 40 m/s from 1.5m height.

Part a: Time to hit ground?

> h = 0.5*g*t² → 1.5 = 4.9*t²
→ t² = 1.5 / 4.9 ≈ 0.306
→ t ≈ 0.553 s

✔ Part a: ~0.55 s

Part b: Horizontal distance?

d = v_x * t = 40 * 0.553 ≈ 22.12 meters

✔ Part b: ~22.1 meters

Part c: Impact velocity?

v_x = 40 m/s
v_y = g*t = 9.8 * 0.553 ≈ 5.42 m/s down

Speed = √(40² + 5.42²) = √(1600 + 29.38) = √1629.38 ≈ 40.37 m/s

Angle = tan⁻¹(5.42 / 40) ≈ tan⁻¹(0.1355) ≈ 7.7° below horizontal

✔ Part c: ~40.4 m/s at ~7.7° below horizontal

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Problem 6:

Cliff diver must clear rocks extending 15m out, cliff is 40m high.

Part a: Time to hit water?

> h = 0.5*g*t² → 40 = 4.9*t²
→ t² = 40 / 4.9 ≈ 8.163
→ t ≈ √8.163 ≈ 2.857 seconds

✔ Part a: ~2.86 s

Part b: Minimum horizontal speed to clear 15m?

Distance = v_x * t → 15 = v_x * 2.857
→ v_x = 15 / 2.857 ≈ 5.25 m/s

✔ Part b: ~5.25 m/s

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Final Answers:

1a. 49.5 meters
1b. 31.4 m/s at 50.5° below horizontal
2a. 0.54 seconds
2b. 3.69 m/s
2c. 6.47 m/s at 55.2° below horizontal
3. 5.53 meters
4. 9.52 kilometers
5a. 0.55 seconds
5b. 22.1 meters
5c. 40.4 m/s at 7.7° below horizontal
6a. 2.86 seconds
6b. 5.25 m/s