A 2 kg object is dropped from a height of 20 m. What is its speed just before it

A 2 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)

A 2 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)

Step 1: Identify the mass of the object (m = 2 kg) and the height from which it is dropped (h = 20 m).
Step 2: Use the acceleration due to gravity (g = 9.8 m/s²).
Step 3: Write the formula for potential energy (PE) at the height: PE = mgh.
Step 4: Write the formula for kinetic energy (KE) just before hitting the ground: KE = 0.5 mv².
Step 5: Set the potential energy equal to the kinetic energy: mgh = 0.5 mv².
Step 6: Notice that the mass (m) cancels out from both sides of the equation.
Step 7: Rearrange the equation to solve for v: v² = 2gh.
Step 8: Substitute the values of g and h into the equation: v² = 2 × 9.8 m/s² × 20 m.
Step 9: Calculate the right side: v² = 392 m²/s².
Step 10: Take the square root to find v: v = √392 m²/s².
Step 11: Calculate the final speed: v ≈ 19.8 m/s.

Conservation of Energy – The principle that the total energy in a closed system remains constant, allowing potential energy to convert into kinetic energy.
Kinematic Equations – Equations that describe the motion of objects under constant acceleration, which can also be used to derive speed from height.