A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is `10 ms^-1` , then the maximum height attained by the stone is (`g= 10 ms^(-2)`)

Options

A. 25 m
B. 10 m
C. 15 m
D. 20 m

Explanation

To solve the problem of finding the maximum height attained by a stone thrown vertically upwards, we can follow these steps:

### Step 1: Understand the Problem
We know that when the stone is at a height equal to half of its maximum height, its speed is given as \(10 \, \text{m/s}\). We need to find the maximum height \(H\) that the stone reaches.

### Step 2: Define Variables
- Let the maximum height be \(H\).
- The height at which the speed is \(10 \, \text{m/s}\) is \(\frac{H}{2}\).
- The acceleration due to gravity, \(g = 10 \, \text{m/s}^2\).

### Step 3: Use the Equation of Motion
We will use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \(v\) = final velocity = \(10 \, \text{m/s}\)
- \(u\) = initial velocity (which we need to find)
- \(a\) = acceleration = \(-g = -10 \, \text{m/s}^2\) (negative because it is acting downwards)
- \(s\) = displacement = \(\frac{H}{2}\)

Substituting the known values into the equation:
\[ (10)^2 = u^2 - 2 \cdot 10 \cdot \frac{H}{2} \]
This simplifies to:
\[ 100 = u^2 - 10H \]
Rearranging gives us:
\[ u^2 = 100 + 10H \quad \text{(Equation 1)} \]

### Step 4: Relate Initial Velocity to Maximum Height
At the maximum height \(H\), the final velocity is \(0\). Using the same kinematic equation:
\[ 0 = u^2 - 2gH \]
Substituting \(g = 10 \, \text{m/s}^2\):
\[ 0 = u^2 - 20H \]
This gives us:
\[ u^2 = 20H \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
Now we can substitute \(u^2\) from Equation 2 into Equation 1:
\[ 20H = 100 + 10H \]
Rearranging gives:
\[ 20H - 10H = 100 \]
\[ 10H = 100 \]
\[ H = 10 \, \text{meters} \]

### Conclusion
The maximum height attained by the stone is \(10 \, \text{meters}\).

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