A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the max

₹0.0

Question: A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)

Options:

Correct Answer: 10 m

Solution:

Using energy conservation, initial kinetic energy = mgh. 0.5 × 1 kg × (10 m/s)² = 1 kg × 10 m/s² × h. h = 5 m.

A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)

A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)

Correct Answer: 5 m

Step 1: Identify the mass of the ball, which is 1 kg.
Step 2: Identify the initial speed of the ball, which is 10 m/s.
Step 3: Identify the acceleration due to gravity, which is 10 m/s².
Step 4: Calculate the initial kinetic energy of the ball using the formula: KE = 0.5 × mass × (speed)².
Step 5: Substitute the values: KE = 0.5 × 1 kg × (10 m/s)² = 0.5 × 1 kg × 100 = 50 Joules.
Step 6: Use the energy conservation principle, where the initial kinetic energy equals the potential energy at maximum height: KE = mgh.
Step 7: Substitute the values into the equation: 50 Joules = 1 kg × 10 m/s² × h.
Step 8: Rearrange the equation to solve for h: h = 50 Joules / (1 kg × 10 m/s²).
Step 9: Calculate h: h = 50 / 10 = 5 meters.

Kinetic and Potential Energy – The problem tests the understanding of the conversion between kinetic energy (when the ball is thrown) and potential energy (at the maximum height).
Energy Conservation Principle – It assesses the application of the principle of conservation of mechanical energy in a vertical motion scenario.
Gravitational Acceleration – The question involves the effect of gravitational acceleration on the motion of the ball.