A 2 kg object is dropped from a height of 15 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
Practice Questions
1 question
Q1
A 2 kg object is dropped from a height of 15 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
17.15 m/s
12.25 m/s
14.14 m/s
10.0 m/s
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 15) = 17.15 m/s.
Questions & Step-by-step Solutions
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Q
Q: A 2 kg object is dropped from a height of 15 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 the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 15) = 17.15 m/s.
Steps: 10
Step 1: Identify the mass of the object (m = 2 kg) and the height from which it is dropped (h = 15 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 at the height will convert to kinetic energy just before it hits the ground.
Step 4: Write the equation for potential energy (PE = mgh) and kinetic energy (KE = 0.5mv²).
Step 5: Set the potential energy equal to the kinetic energy: mgh = 0.5mv².
Step 6: Notice that the mass (m) cancels out from both sides of the equation.
Step 7: Rearrange the equation to solve for v (the speed just before hitting the ground): v² = 2gh.
Step 8: Substitute the values of g and h into the equation: v² = 2 * 9.8 * 15.
Step 9: Calculate the right side: v² = 294.
Step 10: Take the square root of 294 to find v: v = sqrt(294) ≈ 17.15 m/s.
