A 10 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
Practice Questions
1 question
Q1
A 10 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
14 m/s
19.8 m/s
20 m/s
28 m/s
Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.
Questions & Step-by-step Solutions
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Q
Q: A 10 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
Solution: Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.
Steps: 13
Step 1: Identify the mass of the object, which is 10 kg.
Step 2: Identify the height from which the object is dropped, which is 20 m.
Step 3: Identify the acceleration due to gravity, which is 9.8 m/s².
Step 4: Use the formula for potential energy (PE) at the height: PE = mgh, where m is mass, g is gravity, and h is height.
Step 5: Calculate the potential energy: PE = 10 kg × 9.8 m/s² × 20 m.
Step 6: Use the formula for kinetic energy (KE) just before hitting the ground: KE = 0.5mv².
Step 7: Set the potential energy equal to the kinetic energy: mgh = 0.5mv².
Step 8: Cancel the mass (m) from both sides of the equation since it is the same on both sides.
Step 9: Rearrange the equation to solve for v: v² = 2gh.
Step 10: Substitute the values of g and h into the equation: v² = 2 × 9.8 m/s² × 20 m.
Step 11: Calculate the value: v² = 392 m²/s².
Step 12: Take the square root to find v: v = √392 m²/s².
