A stone is thrown in a vertically upward direction with a velocity of `5 m//s`. If the acceleration of the stone during its motion is `10 m//s^(2)` in the downward direction, what will be the height attained by the stone and how much time will take to reach there ?

Options

A. 2 m , 20 sec
B. 1.25 m , 0.5 sec
C. 12.5 m , 5 sec
D. 20 m, 2 sec

Explanation

To solve the problem, we will use the equations of motion under uniform acceleration. Here are the steps to find the height attained by the stone and the time taken to reach that height.

### Step 1: Identify the given values
- Initial velocity (u) = 5 m/s (upward)
- Final velocity (v) = 0 m/s (at the highest point)
- Acceleration (a) = -10 m/s² (downward, hence negative)

### Step 2: Use the third equation of motion to find the height (h)
The third equation of motion is given by:
\[ v^2 = u^2 + 2a s \]
Where:
- \( s \) is the displacement (height in this case).

Rearranging the equation to solve for \( s \):
\[ s = \frac{v^2 - u^2}{2a} \]

Substituting the known values:
\[ s = \frac{0^2 - (5)^2}{2 \times (-10)} \]
\[ s = \frac{-25}{-20} \]
\[ s = \frac{25}{20} \]
\[ s = 1.25 \text{ m} \]

### Step 3: Calculate the time taken to reach the maximum height
We will use the first equation of motion:
\[ v = u + at \]

Rearranging to solve for time \( t \):
\[ t = \frac{v - u}{a} \]

Substituting the known values:
\[ t = \frac{0 - 5}{-10} \]
\[ t = \frac{-5}{-10} \]
\[ t = 0.5 \text{ seconds} \]

### Final Answers
- Height attained by the stone (h) = 1.25 m
- Time taken to reach the maximum height (t) = 0.5 seconds

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