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

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

Solution: Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (15)²/(2*9.8) = 11.5 m.

Steps: 10

Step 1: Identify the given values. The mass of the object (m) is 10 kg, the initial speed (v) is 15 m/s, and the acceleration due to gravity (g) is 9.8 m/s².
Step 2: Understand the concept of energy conservation. When the object is thrown upwards, its kinetic energy (KE) at the bottom will convert to potential energy (PE) at the maximum height.
Step 3: Write the formula for kinetic energy: KE = 0.5 * m * v².
Step 4: Write the formula for potential energy: PE = m * g * h, where h is the height we want to find.
Step 5: Set the kinetic energy equal to the potential energy at the maximum height: 0.5 * m * v² = m * g * h.
Step 6: Notice that the mass (m) cancels out from both sides of the equation, simplifying it to: 0.5 * v² = g * h.
Step 7: Rearrange the equation to solve for height (h): h = v² / (2 * g).
Step 8: Substitute the values into the equation: h = (15)² / (2 * 9.8).
Step 9: Calculate (15)² which is 225, and then calculate 2 * 9.8 which is 19.6.
Step 10: Divide 225 by 19.6 to find h: h = 225 / 19.6 ≈ 11.5 m.