A 2 kg object is dropped from a height of 10 m. What is the speed of the object just before it hits the ground? (g = 9.8 m/s²)

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Q: A 2 kg object is dropped from a height of 10 m. What is the speed of the object just before it hits the ground? (g = 9.8 m/s²)

Solution: Using energy conservation, mgh = 0.5mv²; v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 10 m) = 14 m/s.

Steps: 10

Step 1: Identify the mass of the object (m = 2 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 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 m/s² × 10 m.
Step 9: Calculate the right side: v² = 196 m²/s².
Step 10: Take the square root of both sides to find v: v = sqrt(196 m²/s²) = 14 m/s.