Understanding the Problem
When a ball is thrown vertically upwards with an initial speed \( u \), it will ascend until it reaches its peak height before descending. The distance covered during the last \( t \) seconds of its ascent can be calculated using kinematic equations.
Key Concepts
- Initial Velocity (u): The speed at which the ball is thrown upward.
- Acceleration (a): The acceleration due to gravity, which is approximately \( -9.81 \, m/s^2 \) (negative because it acts downward).
- Time (t): The last \( t \) seconds of ascent.
Kinematic Equation
The distance \( s \) covered during the last \( t \) seconds can be calculated using the formula:
- \( s = u*t - \frac{1}{2}gt^2 \)
Where \( g \) is the acceleration due to gravity.
Derivation of Distance
1. Total Time of Ascent (T): The total time taken to reach the maximum height can be found using \( T = \frac{u}{g} \).
2. Distance Covered in \( T-t \) Seconds: The distance covered during the time \( T-t \) can be found using:
- \( s_{T-t} = u(T-t) - \frac{1}{2}g(T-t)^2 \)
3. Distance Covered in \( T \) Seconds: The total distance covered in time \( T \):
- \( s_T = uT - \frac{1}{2}gT^2 \)
4. Distance in Last \( t \) Seconds: The difference gives:
- \( s = s_T - s_{T-t} \)
Final Distance Formula
Thus, the distance covered during the last \( t \) seconds of ascent is:
- \( s = ut - \frac{1}{2}g(t^2) \)
This formula effectively captures the distance traveled by the ball during its final moments of ascent before it reaches its peak.