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

Options

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

Explanation

To solve the problem of finding the distance covered during the last \( t \) seconds of ascent when a ball is thrown vertically upwards with an initial speed \( u \), we can follow these steps:

### Step 1: Understand the motion of the ball
When a ball is thrown upwards, it will ascend until it reaches its maximum height, at which point its velocity will be zero. The time taken to reach the maximum height can be calculated using the formula:
\[ v = u - gt \]
where \( v \) is the final velocity (0 at the maximum height), \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time taken to reach that height.

### Step 2: Calculate the total time of ascent
Setting \( v = 0 \):
\[ 0 = u - gt_{max} \]
From this, we can find the time to reach maximum height:
\[ t_{max} = \frac{u}{g} \]

### Step 3: Determine the time interval for the last \( t \) seconds of ascent
The last \( t \) seconds of ascent means we are looking at the time interval from \( t_{max} - t \) to \( t_{max} \).

### Step 4: Calculate the height reached at \( t_{max} \)
The height \( H \) reached at the maximum height can be calculated using:
\[ H = ut_{max} - \frac{1}{2} g t_{max}^2 \]
Substituting \( t_{max} = \frac{u}{g} \):
\[ H = u \left(\frac{u}{g}\right) - \frac{1}{2} g \left(\frac{u}{g}\right)^2 \]
\[ H = \frac{u^2}{g} - \frac{1}{2} \frac{u^2}{g} = \frac{u^2}{2g} \]

### Step 5: Calculate the height at \( t_{max} - t \)
Now, we need to find the height at \( t_{max} - t \):
\[ H_{t_{max} - t} = u(t_{max} - t) - \frac{1}{2} g (t_{max} - t)^2 \]
Substituting \( t_{max} = \frac{u}{g} \):
\[ H_{t_{max} - t} = u\left(\frac{u}{g} - t\right) - \frac{1}{2} g \left(\frac{u}{g} - t\right)^2 \]

### Step 6: Calculate the distance covered during the last \( t \) seconds
The distance covered during the last \( t \) seconds is the difference in heights:
\[ \text{Distance} = H_{t_{max}} - H_{t_{max} - t} \]
Substituting the values we calculated:
\[ \text{Distance} = H - H_{t_{max} - t} \]

### Final Result
After simplifying the calculations, we find that the distance covered during the last \( t \) seconds of ascent is given by:
\[ \text{Distance} = ut - \frac{1}{2} g t^2 \]