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? (g = 9.8 m/s²)

Options:

Correct Answer: 30.4 m

Solution:

Using conservation of energy, KE at the bottom = PE at the maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2 * 9.8) = 20.4 m.

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)

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.