A 1 kg ball is thrown vertically upwards with a speed of 20 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 20 m/s. What is the maximum height it reaches?

Solution: Using energy conservation: KE_initial = PE_max; 0.5 × m × v² = mgh; h = v²/(2g) = (20 m/s)²/(2 × 9.8 m/s²) = 20.4 m.

Steps: 9

Step 1: Identify the initial speed of the ball, which is 20 m/s.
Step 2: Recognize that the ball's kinetic energy (KE) will convert to potential energy (PE) at the maximum height.
Step 3: Write the formula for kinetic energy: KE_initial = 0.5 × m × v².
Step 4: Write the formula for potential energy at maximum height: PE_max = mgh.
Step 5: Set the initial kinetic energy equal to the maximum potential energy: 0.5 × m × v² = mgh.
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 / (2g).
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 m²/s² / 19.6 m/s² = 20.4 m.