A 3 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the max

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Question: A 3 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

Options:

Correct Answer: 5.1 m

Solution:

Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (10)²/(2*9.8) = 5.1 m.

A 3 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A 3 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

Correct Answer: 5.1 m

Step 1: Identify the mass of the ball, which is 3 kg, and the initial speed, which is 10 m/s.
Step 2: Recognize 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 velocity.
Step 4: Calculate the kinetic energy at the bottom: KE = 0.5 * 3 kg * (10 m/s)².
Step 5: Calculate KE: KE = 0.5 * 3 * 100 = 150 Joules.
Step 6: Write the formula for potential energy at the maximum height: PE = m * g * h, where g is the acceleration due to gravity (9.8 m/s²) and h is the height.
Step 7: Set the kinetic energy equal to the potential energy at the maximum height: 150 Joules = 3 kg * 9.8 m/s² * h.
Step 8: Rearrange the equation to solve for h: h = 150 / (3 * 9.8).
Step 9: Calculate h: h = 150 / 29.4 ≈ 5.1 m.

Conservation of Energy – The principle that the total energy in a closed system remains constant, allowing the conversion of kinetic energy to potential energy in this scenario.
Kinematic Equations – Understanding the relationship between velocity, acceleration, and displacement in motion, particularly in vertical motion under gravity.
Gravitational Potential Energy – The energy an object possesses due to its position in a gravitational field, calculated as PE = mgh.