This worksheet provides answers and detailed solutions to common free fall physics problems. Understanding free fall is crucial for grasping fundamental concepts in Newtonian mechanics. Remember, in idealized free fall problems, we ignore air resistance. Gravity is considered the only force acting on the object.
Key Concepts to Remember:
- Acceleration due to gravity (g): Approximately 9.8 m/s² on Earth, directed downwards. This means velocity changes by 9.8 m/s every second.
- Kinematic Equations: These equations relate displacement (Δy), initial velocity (v₀), final velocity (v), acceleration (a = g), and time (t). The most commonly used are:
- Δy = v₀t + (1/2)at²
- v = v₀ + at
- v² = v₀² + 2aΔy
Problem Set & Solutions
Problem 1: A ball is dropped from rest from a height of 20 meters. How long does it take to hit the ground?
Solution:
- Known: Δy = -20 m (negative because downward), v₀ = 0 m/s, a = -9.8 m/s²
- Unknown: t
- Equation: Δy = v₀t + (1/2)at²
- Solution: -20 = 0*t + (1/2)(-9.8)t² => t² = 40/9.8 => t ≈ 2.02 seconds
Problem 2: A stone is thrown vertically upward with an initial velocity of 15 m/s. What is its maximum height?
Solution:
- Known: v₀ = 15 m/s, v = 0 m/s (at maximum height, velocity is momentarily zero), a = -9.8 m/s²
- Unknown: Δy
- Equation: v² = v₀² + 2aΔy
- Solution: 0 = 15² + 2(-9.8)Δy => Δy = 225 / (19.6) ≈ 11.48 meters
Problem 3: An object is dropped from a cliff and takes 3 seconds to hit the ground. How high is the cliff?
Solution:
- Known: t = 3 s, v₀ = 0 m/s, a = -9.8 m/s²
- Unknown: Δy
- Equation: Δy = v₀t + (1/2)at²
- Solution: Δy = 0*(3) + (1/2)(-9.8)(3)² = -44.1 m. The height of the cliff is 44.1 meters.
Problem 4: A ball is thrown vertically upward at 20 m/s. Find its velocity after 2 seconds.
Solution:
- Known: v₀ = 20 m/s, a = -9.8 m/s², t = 2 s
- Unknown: v
- Equation: v = v₀ + at
- Solution: v = 20 + (-9.8)(2) = 0.4 m/s (upward, since it's positive).
Problem 5: A rock is thrown downward from a bridge with an initial velocity of 5 m/s. It hits the water below 2 seconds later. How high is the bridge above the water?
Solution:
- Known: v₀ = -5 m/s (negative since downward), a = -9.8 m/s², t = 2s
- Unknown: Δy
- Equation: Δy = v₀t + (1/2)at²
- Solution: Δy = (-5)(2) + (1/2)(-9.8)(2)² = -10 -19.6 = -29.6 m. The height of the bridge is 29.6 meters.
This worksheet covers basic free-fall problems. More complex scenarios might involve projectiles launched at angles, requiring vector decomposition of velocity and separate consideration of horizontal and vertical motion. Always remember to carefully define your coordinate system (e.g., positive upwards or downwards) to avoid sign errors.
