A 2 kg ball is dropped from a height of 10 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)
Practice Questions
1 question
Q1
A 2 kg ball is dropped from a height of 10 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)
10 m/s
14 m/s
20 m/s
15 m/s
Using conservation of energy, Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 1/2 mv^2. Solving gives v = sqrt(2gh) = sqrt(2 * 10 m/s² * 10 m) = 14.14 m/s.
Questions & Step-by-step Solutions
1 item
Q
Q: A 2 kg ball is dropped from a height of 10 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)
Solution: Using conservation of energy, Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 1/2 mv^2. Solving gives v = sqrt(2gh) = sqrt(2 * 10 m/s² * 10 m) = 14.14 m/s.
Steps: 9
Step 1: Identify the mass of the ball, which is 2 kg, and the height from which it is dropped, which is 10 m.
Step 2: Understand that when the ball is dropped, it has potential energy due to its height.
Step 3: Use the formula for potential energy (PE) at height: PE = mgh, where m is mass, g is acceleration due to gravity (10 m/s²), and h is height (10 m).
Step 4: Calculate the potential energy: PE = 2 kg * 10 m/s² * 10 m = 200 Joules.
Step 5: Know that just before the ball hits the ground, all potential energy converts to kinetic energy (KE).
Step 6: Use the formula for kinetic energy: KE = 1/2 mv².
Step 7: Set the potential energy equal to the kinetic energy: 200 Joules = 1/2 * 2 kg * v².
Step 8: Simplify the equation: 200 = 1 kg * v², which means v² = 200.
Step 9: Take the square root of both sides to find v: v = sqrt(200) = sqrt(2 * 100) = sqrt(2) * 10 = 14.14 m/s.
