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

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Question: A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?

Options:

Correct Answer: 20 m

Solution:

Using conservation of energy, KE at launch = PE at max height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2*9.8) = 20.41 m.

A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?

A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?

Correct Answer: 20.41 m

Step 1: Identify the mass of the ball (m = 2 kg) and the initial speed (v = 20 m/s).
Step 2: Understand that when the ball is thrown upwards, it has kinetic energy (KE) at the start and potential energy (PE) at the maximum height.
Step 3: Write the formula for kinetic energy: KE = 0.5 * m * v².
Step 4: Write the formula 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 5: Set the kinetic energy 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.
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 = 20.41 m.

Conservation of Energy – The principle that the total energy in a closed system remains constant, allowing the conversion of kinetic energy (KE) to potential energy (PE) as the ball rises.
Kinetic Energy – The energy possessed by an object due to its motion, calculated as KE = 0.5mv².
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, which affects the ball's ascent and descent.