A 2 kg ball is dropped from a height of 20 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 20 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)
10 m/s
20 m/s
15 m/s
5 m/s
Using energy conservation, Potential Energy = Kinetic Energy at the ground: mgh = 0.5 mv^2; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 20 m) = 20 m/s
Questions & Step-by-step Solutions
1 item
Q
Q: A 2 kg ball is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)
Solution: Using energy conservation, Potential Energy = Kinetic Energy at the ground: mgh = 0.5 mv^2; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 20 m) = 20 m/s
Steps: 10
Step 1: Identify the mass of the ball, which is 2 kg.
Step 2: Identify the height from which the ball is dropped, which is 20 m.
Step 3: Identify the acceleration due to gravity, which is given as 10 m/s².
Step 4: Use the formula for potential energy (PE) at the height: PE = mgh, where m is mass, g is gravity, and h is height.
Step 5: Calculate the potential energy: PE = 2 kg * 10 m/s² * 20 m = 400 Joules.
Step 6: When the ball falls, this potential energy converts to kinetic energy (KE) just before it hits the ground.
Step 7: Use the formula for kinetic energy: KE = 0.5 * m * v², where v is the speed just before hitting the ground.
Step 8: Set the potential energy equal to the kinetic energy: 400 Joules = 0.5 * 2 kg * v².
Step 9: Simplify the equation: 400 = 1 * v², so v² = 400.
Step 10: Take the square root of both sides to find v: v = sqrt(400) = 20 m/s.
