A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the max

A 1 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 1 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 1 kg.
Step 2: Identify the initial speed of the ball, which is 20 m/s.
Step 3: Identify the acceleration due to gravity, which is 9.8 m/s².
Step 4: Calculate the initial kinetic energy of the ball using the formula: KE = 0.5 × mass × (speed)².
Step 5: Plug in the values: KE = 0.5 × 1 kg × (20 m/s)² = 0.5 × 1 × 400 = 200 Joules.
Step 6: Use the potential energy formula at maximum height: PE = mass × g × height (h).
Step 7: Set the initial kinetic energy equal to the potential energy at maximum height: 200 Joules = 1 kg × 9.8 m/s² × h.
Step 8: Rearrange the equation to solve for height (h): h = 200 Joules / (1 kg × 9.8 m/s²).
Step 9: Calculate h: h = 200 / 9.8 ≈ 20.4 meters.

Kinetic and Potential Energy – The question tests the understanding of the conversion between kinetic energy (when the ball is thrown) and potential energy (at maximum height).
Energy Conservation – It assesses the ability to apply the principle of conservation of mechanical energy in a vertical motion scenario.
Gravitational Acceleration – The problem requires knowledge of gravitational acceleration and its effect on an object's motion.