A 2 kg object is thrown vertically upwards with a speed of 10 m/s. What is the m

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Question: A 2 kg object 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: KE at bottom = PE at top; 1/2 mv^2 = mgh; h = v^2/(2g) = (10^2)/(2*10) = 5 m

A 2 kg object is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)

A 2 kg object is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)

Step 1: Identify the mass of the object (m = 2 kg) and the initial speed (v = 10 m/s).
Step 2: Recognize that when the object is thrown upwards, it has kinetic energy (KE) at the bottom and potential energy (PE) at the maximum height.
Step 3: Write the formula for kinetic energy: KE = 1/2 * m * v^2.
Step 4: Write the formula for potential energy: PE = m * g * h, where g = 10 m/s² and h is the height we want to find.
Step 5: Set the kinetic energy equal to the potential energy at the maximum height: 1/2 * m * v^2 = m * g * h.
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 = v^2 / (2 * g).
Step 8: Substitute the values into the equation: h = (10^2) / (2 * 10).
Step 9: Calculate the height: h = 100 / 20 = 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 an object moving vertically under the influence of gravity.
Energy Conservation Principle – It assesses the ability to apply the principle of conservation of energy, where the kinetic energy at the point of launch is converted into potential energy at the maximum height.
Gravitational Acceleration – The problem involves understanding the effect of gravitational acceleration on the motion of the object.