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

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

Solution: Using energy conservation: KE_initial = PE_max; 1/2 * m * v^2 = m * g * h; h = v^2 / (2g) = (15^2) / (2 * 10) = 11.25 m

Steps: 10

Step 1: Identify the mass of the object (m = 10 kg) and the initial speed (v = 15 m/s).
Step 2: Recognize that when the object is thrown upwards, it has kinetic energy (KE) that will convert to potential energy (PE) at the maximum height.
Step 3: Write the formula for kinetic energy: KE_initial = 1/2 * m * v^2.
Step 4: Write the formula for potential energy at maximum height: PE_max = 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 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, simplifying it to: 1/2 * v^2 = g * h.
Step 7: Rearrange the equation to solve for height (h): h = v^2 / (2g).
Step 8: Substitute the values into the equation: h = (15^2) / (2 * 10).
Step 9: Calculate 15^2, which is 225, and then calculate 2 * 10, which is 20.
Step 10: Divide 225 by 20 to find h: h = 225 / 20 = 11.25 m.