A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?

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Q: A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?

Solution: Using energy conservation, mgh = 0.5 mv², h = v²/(2g) = (10 m/s)²/(2 × 9.8 m/s²) = 5.1 m.

Steps: 10

Step 1: Identify the mass of the ball, which is 1 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 initial 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: 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.