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)

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Q: 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)

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

Steps: 9

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.