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A ball is thrown vertically upward with speed u, the distance coverd during the last t second of it's ascent is?
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A ball is thrown vertically upward with speed u, the distance coverd d...
Understanding the Problem
When a ball is thrown vertically upward with an initial speed "u," it experiences deceleration due to gravity. The key objective is to find the distance covered during the last "t" seconds of its ascent.
Key Concepts
- Initial Speed (u): The speed at which the ball is thrown upward.
- Acceleration due to Gravity (g): The constant acceleration acting downward, approximately 9.81 m/s².
- Time of Ascent (T): The total time taken to reach the maximum height. This can be calculated using the formula: T = u/g.
Distance Covered in the Last t Seconds
To find the distance covered during the last t seconds of ascent, we will use the following steps:
1. Find the Velocity at Time T-t:
- The velocity at any time "t" during the ascent can be calculated as:
- V = u - g*t
2. Distance Covered in Last t Seconds:
- The distance "s" covered in the last t seconds can be derived using:
- s = V_initial * t + 0.5 * (-g) * t²
- Here, V_initial represents the velocity at time T - t, and the negative sign indicates the direction of gravity.
Final Formula
After substituting the values, the distance covered during the last t seconds of ascent can be expressed as:
- s = (u - g*(T - t)) * t - 0.5 * g * t²
This formula allows you to calculate the distance covered during the last t seconds of ascent effectively.
Conclusion
Understanding these principles ensures a comprehensive grasp of projectile motion, particularly in vertical throws.
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A ball is thrown vertically upward with speed u, the distance coverd during the last t second of it's ascent is?
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