A 10 kg object is dropped from a height of 5 m. What is its speed just before it

A 10 kg object is dropped from a height of 5 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

A 10 kg object is dropped from a height of 5 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

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 5 m.
Step 3: Use the acceleration due to gravity, which is given as g = 9.8 m/s².
Step 4: 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 5: Write the formula for potential energy (PE) at the height: PE = mgh, where m is mass, g is gravity, and h is height.
Step 6: Write the formula for kinetic energy (KE) just before hitting the ground: KE = (1/2)mv², where v is the speed.
Step 7: Set the potential energy equal to the kinetic energy: mgh = (1/2)mv².
Step 8: Notice that the mass (m) cancels out from both sides of the equation.
Step 9: Rearrange the equation to solve for v: v² = 2gh.
Step 10: Substitute the values for g and h into the equation: v² = 2 × 9.8 m/s² × 5 m.
Step 11: Calculate the right side: v² = 98 m²/s².
Step 12: Take the square root of both sides to find v: v = sqrt(98 m²/s²).
Step 13: Calculate the square root: v ≈ 9.9 m/s, which can be rounded to approximately 10 m/s.

Energy Conservation – The principle that the total energy in a closed system remains constant, allowing potential energy to convert into kinetic energy.
Kinematics – The study of motion, particularly how objects move under the influence of gravity.
Gravitational Acceleration – The acceleration due to gravity, which is approximately 9.8 m/s² near the Earth's surface.