A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
Practice Questions
1 question
Q1
A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
20.4 m
30.4 m
40.4 m
50.4 m
Using conservation of energy, KE at the bottom = PE at the maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2 * 9.8) = 20.4 m.
Questions & Step-by-step Solutions
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Q
Q: A 2 kg ball is thrown vertically upwards with a speed of 20 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 maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2 * 9.8) = 20.4 m.
Steps: 9
Step 1: Identify the mass of the ball, which is 2 kg, and the initial speed, which is 20 m/s.
Step 2: Understand that when the ball 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 = 0.5 * m * v², where m is mass and v is speed.
Step 4: Write the formula for potential energy: PE = m * g * h, where g is the acceleration due to gravity (9.8 m/s²) and h is the height.
Step 5: Set the kinetic energy at the bottom 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 = (20 m/s)² / (2 * 9.8 m/s²).
Step 9: Calculate the height: h = 400 / 19.6 = 20.4 m.
