A 2 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the max

Practice Questions

Q1
A 2 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?
  1. 5 m
  2. 10 m
  3. 15 m
  4. 20 m

Questions & Step-by-Step Solutions

A 2 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?
Correct Answer: 5.1 m
  • Step 1: Identify the mass of the ball, which is 2 kg, and the initial speed, which is 10 m/s.
  • Step 2: Understand that when the ball is thrown upwards, it will rise until it stops momentarily at the maximum height.
  • Step 3: Use the principle of energy conservation, which states that the kinetic energy (energy of motion) will convert to potential energy (energy of position) at the maximum height.
  • Step 4: Write the equation for kinetic energy: KE = 0.5 * m * v², where m is mass and v is speed.
  • Step 5: Write the equation for potential energy: PE = m * g * h, where g is the acceleration due to gravity (approximately 9.8 m/s²) and h is the height.
  • Step 6: Set the kinetic energy equal to the potential energy at the maximum height: 0.5 * m * v² = m * g * h.
  • Step 7: Notice that the mass (m) cancels out from both sides of the equation, simplifying it to: 0.5 * v² = g * h.
  • Step 8: Rearrange the equation to solve for height (h): h = v² / (2 * g).
  • Step 9: Substitute the values into the equation: h = (10 m/s)² / (2 * 9.8 m/s²).
  • Step 10: Calculate the height: h = 100 m²/s² / 19.6 m/s² = 5.1 m.
  • Energy Conservation – The principle that the total mechanical energy (kinetic + potential) in a closed system remains constant if only conservative forces are acting.
  • Kinematics – The study of motion without considering the forces that cause it, particularly the equations of motion for objects under constant acceleration.
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