A 5 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)
Practice Questions
1 question
Q1
A 5 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)
14 m/s
20 m/s
10 m/s
15 m/s
Using energy conservation, Potential Energy = Kinetic Energy. mgh = 1/2 mv². v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s.
Questions & Step-by-step Solutions
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Q
Q: A 5 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)
Solution: Using energy conservation, Potential Energy = Kinetic Energy. mgh = 1/2 mv². v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s.
Steps: 12
Step 1: Identify the mass of the object (m = 5 kg) and the height from which it is dropped (h = 10 m).
Step 2: Use the acceleration due to gravity (g = 9.8 m/s²).
Step 3: Understand that when the object is dropped, its potential energy (PE) at the height will convert to kinetic energy (KE) just before it hits the ground.
Step 4: Write the formula for potential energy: PE = mgh.
Step 5: Write the formula for kinetic energy: KE = 1/2 mv².
Step 6: Set the potential energy equal to the kinetic energy: mgh = 1/2 mv².
Step 7: Notice that the mass (m) cancels out from both sides of the equation.
Step 8: Rearrange the equation to solve for v (speed): v² = 2gh.
Step 9: Substitute the values of g and h into the equation: v² = 2 × 9.8 m/s² × 10 m.
Step 10: Calculate the right side: v² = 196 m²/s².
Step 11: Take the square root of both sides to find v: v = √196 m²/s².
