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A 1 kg ball is thrown upwards with a speed of 20 m/s. What is the maximum height
A 1 kg ball is thrown upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²) (2021)
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A 1 kg ball is thrown upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²) (2021)
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Maximum Height (h) = v^2 / (2g) = (20 m/s)^2 / (2 * 10 m/s²) = 20 m
Questions & Step-by-step Solutions
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Q: A 1 kg ball is thrown upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²) (2021)
Solution:
Maximum Height (h) = v^2 / (2g) = (20 m/s)^2 / (2 * 10 m/s²) = 20 m
Steps: 8
Show Steps
Step 1: Identify the initial speed (v) of the ball, 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) reached by the ball: h = v^2 / (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: h = 400 m²/s² / 20 m/s² = 20 m.
Step 8: Conclude that the maximum height reached by the ball is 20 meters.
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