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

A 2 kg ball is thrown vertically upwards with a velocity of 20 m/s. What is the maximum height reached? (2020)

A 2 kg ball is thrown vertically upwards with a velocity of 20 m/s. What is the maximum height reached? (2020)

Step 1: Identify the initial velocity (u) of the ball, which is 20 m/s.
Step 2: Recognize that the final velocity (v) at the maximum height is 0 m/s because the ball stops rising at that point.
Step 3: Note the acceleration (a) due to gravity, which is -9.8 m/s² (negative because it acts downwards).
Step 4: Use the formula v² = u² + 2as to find the maximum height (h).
Step 5: Substitute the values into the formula: 0 = (20 m/s)² + 2(-9.8 m/s²)(h).
Step 6: Simplify the equation: 0 = 400 - 19.6h.
Step 7: Rearrange the equation to solve for h: 19.6h = 400.
Step 8: Divide both sides by 19.6 to find h: h = 400 / 19.6.
Step 9: Calculate the value: h ≈ 20.4 m.

Kinematics – The question tests the understanding of kinematic equations, specifically how to apply them to calculate maximum height in projectile motion.
Energy Conservation – It also touches on the concept of energy conservation, where kinetic energy is converted to potential energy at the maximum height.