A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the max

Practice Questions

Q1
A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
  1. 20.4 m
  2. 30.4 m
  3. 40.4 m
  4. 50.4 m

Questions & Step-by-Step Solutions

A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
  • Step 1: Identify the mass of the ball, which is 2 kg, and the initial speed, which is 20 m/s.
  • Step 2: Understand that when the ball is thrown upwards, it has kinetic energy (KE) at the bottom and potential energy (PE) at the maximum height.
  • Step 3: Write the formula for kinetic energy: KE = 0.5 * m * v², where m is mass and v is speed.
  • Step 4: Write the formula for potential energy: PE = m * g * h, where g is the acceleration due to gravity (9.8 m/s²) and h is the height.
  • Step 5: Set the kinetic energy at the bottom equal to the potential energy at the maximum height: 0.5 * m * v² = m * g * h.
  • Step 6: Notice that the mass (m) cancels out from both sides of the equation, simplifying it to: 0.5 * v² = g * h.
  • Step 7: Rearrange the equation to solve for height (h): h = v² / (2 * g).
  • Step 8: Substitute the values into the equation: h = (20 m/s)² / (2 * 9.8 m/s²).
  • Step 9: Calculate the height: h = 400 / 19.6 = 20.4 m.
  • Conservation of Energy – The principle that the total energy in a closed system remains constant; in this case, kinetic energy is converted to potential energy.
  • Kinetic Energy (KE) – The energy possessed by an object due to its motion, calculated as KE = 0.5mv².
  • Potential Energy (PE) – The energy stored in an object due to its height above the ground, calculated as PE = mgh.
  • Maximum Height Calculation – Determining the height reached by an object when its velocity becomes zero at the peak of its trajectory.
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