A 10 kg object is thrown upwards with a velocity of 15 m/s. What is the maximum

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Question: A 10 kg object is thrown upwards with a velocity of 15 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

Options:

Correct Answer: 11.5 m

Solution:

Using energy conservation: KE_initial = PE_max; 0.5 * m * v² = mgh; h = v² / (2g) = (15 m/s)² / (2 * 9.8 m/s²) = 11.5 m.

A 10 kg object is thrown upwards with a velocity of 15 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A 10 kg object is thrown upwards with a velocity of 15 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

Step 1: Identify the mass of the object (m = 10 kg) and the initial velocity (v = 15 m/s).
Step 2: Recognize that when the object reaches its maximum height, all its initial kinetic energy will be converted into potential energy.
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 kinetic energy equal to the 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 = (15 m/s)² / (2 * 9.8 m/s²).
Step 9: Calculate (15 m/s)² = 225 m²/s².
Step 10: Calculate 2 * 9.8 m/s² = 19.6 m/s².
Step 11: Divide 225 m²/s² by 19.6 m/s² to find h: h = 225 / 19.6.
Step 12: Calculate the final height: h ≈ 11.5 m.

Kinetic Energy and Potential Energy – The question tests the understanding of the conversion between kinetic energy (KE) and potential energy (PE) in the context of projectile motion.
Energy Conservation Principle – It assesses the ability to apply the principle of conservation of mechanical energy to find the maximum height reached by an object.
Kinematic Equations – The question indirectly tests knowledge of kinematic equations, particularly the relationship between initial velocity, acceleration due to gravity, and maximum height.