A 2 kg ball is thrown vertically upwards with a speed of 15 m/s. What is the max

Practice Questions

Q1
A 2 kg ball is thrown vertically upwards with a speed of 15 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 15 m/s. What is the maximum height it reaches?
Correct Answer: 11.47 m
  • Step 1: Identify the mass of the ball, which is 2 kg, and the initial speed, which is 15 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 top.
  • Step 3: Write the formula for kinetic energy: KE = 0.5 * m * v^2, 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 (approximately 9.8 m/s^2) and h is the height.
  • Step 5: Set the kinetic energy at the bottom equal to the potential energy at the top: 0.5 * m * v^2 = m * g * h.
  • Step 6: Notice that the mass (m) cancels out from both sides of the equation.
  • Step 7: Rearrange the equation to solve for height (h): h = v^2 / (2 * g).
  • Step 8: Substitute the values into the equation: h = (15^2) / (2 * 9.8).
  • Step 9: Calculate 15^2, which is 225.
  • Step 10: Calculate 2 * 9.8, which is 19.6.
  • Step 11: Divide 225 by 19.6 to find h: h = 225 / 19.6.
  • Step 12: Perform the division to find the maximum height, which is approximately 11.47 m.
  • Conservation of Energy – The principle that the total energy in a closed system remains constant; in this case, kinetic energy (KE) is converted to potential energy (PE) as the ball rises.
  • Kinetic Energy – The energy possessed by an object due to its motion, calculated as KE = 0.5mv^2.
  • Potential Energy – The energy stored in an object due to its height above the ground, calculated as PE = mgh.
  • Gravitational Acceleration – The acceleration due to gravity, approximately 9.8 m/s² on the surface of the Earth.
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