A 2 kg ball is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches?
Practice Questions
1 question
Q1
A 2 kg ball is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches?
5 m
10 m
15 m
20 m
Using conservation of energy, KE at the bottom = PE at the top. 0.5mv^2 = mgh. Solving gives h = v^2/(2g) = (15^2)/(2*9.8) = 11.47 m.
Questions & Step-by-step Solutions
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Q
Q: A 2 kg ball is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches?
Solution: Using conservation of energy, KE at the bottom = PE at the top. 0.5mv^2 = mgh. Solving gives h = v^2/(2g) = (15^2)/(2*9.8) = 11.47 m.
Steps: 12
Step 1: Identify the mass of the ball, which is 2 kg, and the initial speed, which is 15 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 top.
Step 3: Write the formula for kinetic energy: KE = 0.5 * m * v^2, 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 (approximately 9.8 m/s^2) and h is the height.
Step 5: Set the kinetic energy at the bottom equal to the potential energy at the top: 0.5 * 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 = (15^2) / (2 * 9.8).
Step 9: Calculate 15^2, which is 225.
Step 10: Calculate 2 * 9.8, which is 19.6.
Step 11: Divide 225 by 19.6 to find h: h = 225 / 19.6.
Step 12: Perform the division to find the maximum height, which is approximately 11.47 m.
