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A ball is dropped from the top of 100m high tower on a planet. in the last ½ second before hitting the ground.it covers a distance of 19m.Accelaration due to gravity near the surface on that planet?
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A ball is dropped from the top of 100m high tower on a planet. in the ...
Understanding the Problem
A ball is dropped from a height of 100 meters, and in the last ½ second before it hits the ground, it covers a distance of 19 meters. We need to find the acceleration due to gravity on that planet.
Key Concepts
- Distance Covered: The last ½ second covers 19 meters.
- Total Height: The total height from which the ball is dropped is 100 meters.
Distance Calculation
To find the acceleration, we can use the equations of motion. The distance covered in the last ½ second can be calculated using:
- Distance = Initial Velocity × Time + 0.5 × Acceleration × Time²
Since the ball is dropped, the initial velocity (u) is 0. We substitute the values:
- Distance = 0 × 0.5 + 0.5 × g × (0.5)² = 0.5 × g × 0.25 = 0.125g
Given the distance covered is 19 meters:
- 0.125g = 19
- g = 19 / 0.125 = 152 m/s²
Acceleration Due to Gravity
Thus, the acceleration due to gravity on that planet is approximately:
- g ≈ 152 m/s²
Conclusion
This high value of gravity indicates that the planet has a significantly stronger gravitational force compared to Earth, where g is approximately 9.81 m/s². The calculation reveals how motion equations can effectively determine gravitational acceleration in various scenarios.
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A ball is dropped from the top of 100m high tower on a planet. in the last ½ second before hitting the ground.it covers a distance of 19m.Accelaration due to gravity near the surface on that planet?
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