An object is thrown upwards with a velocity of 20 m/s. How high will it go befor

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Question: An object is thrown upwards with a velocity of 20 m/s. How high will it go before coming to rest? (g = 10 m/s²)

Options:

Correct Answer: 40 m

Solution:

Using the formula h = v²/(2g) = (20 m/s)² / (2 * 10 m/s²) = 20 m.

An object is thrown upwards with a velocity of 20 m/s. How high will it go before coming to rest? (g = 10 m/s²)

An object is thrown upwards with a velocity of 20 m/s. How high will it go before coming to rest? (g = 10 m/s²)

Step 1: Identify the initial velocity (v) of the object, which is 20 m/s.
Step 2: Identify the acceleration due to gravity (g), which is 10 m/s².
Step 3: Use the formula for maximum height (h) when an object is thrown upwards: h = v² / (2g).
Step 4: Substitute the values into the formula: h = (20 m/s)² / (2 * 10 m/s²).
Step 5: Calculate (20 m/s)², which equals 400 m²/s².
Step 6: Calculate 2 * 10 m/s², which equals 20 m/s².
Step 7: Divide 400 m²/s² by 20 m/s² to find h: 400 m²/s² / 20 m/s² = 20 m.
Step 8: Conclude that the maximum height the object will reach is 20 meters.

Kinematics – The question tests the understanding of the kinematic equations, specifically the relationship between initial velocity, acceleration due to gravity, and maximum height reached by an object in free fall.
Energy Conservation – It also touches on the concept of energy conservation, where kinetic energy is converted to potential energy at the peak of the object's trajectory.