If a ball is thrown vertically upwards with speed `u`, the distance covered during the last `t` second of its ascent is

Options

A. `1/2g t^(2)`
B. `ut-1/2g t^(2)`
C. `(u-g t)t`
D. ut

Explanation

To find the distance covered by a ball thrown vertically upwards during the last `t` seconds of its ascent, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- When a ball is thrown upwards with an initial speed `u`, it will rise until it reaches its maximum height, where its velocity becomes zero. The time taken to reach this maximum height is equal to the time taken to descend back to the original height.

2. **Time of Ascent**:
- The total time of ascent can be calculated using the formula:
\[ t_{total} = \frac{u}{g} \]
- Here, `g` is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)).

3. **Distance Covered in Last t Seconds**:
- To find the distance covered in the last `t` seconds of ascent, we can use the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
- In this case, we need to consider the distance covered in the last `t` seconds. The distance covered in the last `t` seconds can be expressed as:
\[ s_{last} = s(t_{total}) - s(t_{total} - t) \]
- Where \(s(t)\) is the distance covered in time `t`.

4. **Calculating s(t_total)**:
- The distance covered in the total time of ascent \(t_{total}\) is:
\[ s(t_{total}) = u \cdot t_{total} - \frac{1}{2} g t_{total}^2 \]
- Substituting \(t_{total} = \frac{u}{g}\):
\[ s(t_{total}) = u \cdot \frac{u}{g} - \frac{1}{2} g \left(\frac{u}{g}\right)^2 \]
\[ = \frac{u^2}{g} - \frac{1}{2} \cdot \frac{u^2}{g} = \frac{u^2}{2g} \]

5. **Calculating s(t_total - t)**:
- The distance covered in the time \(t_{total} - t\) is:
\[ s(t_{total} - t) = u \cdot (t_{total} - t) - \frac{1}{2} g (t_{total} - t)^2 \]

6. **Final Calculation**:
- The distance covered in the last `t` seconds is:
\[ s_{last} = s(t_{total}) - s(t_{total} - t) \]
- After substituting the values and simplifying, we find that:
\[ s_{last} = u \cdot t - \frac{1}{2} g t^2 \]

### Final Answer:
The distance covered during the last `t` seconds of ascent is:
\[ s_{last} = u \cdot t - \frac{1}{2} g t^2 \]