A stone is dropped from a hill of height 180 m. Two seconds later another stone is dropped from a point P below the top of the hill . If the two stones reach the groud simultaneously, the height of P from the ground is `(g = 10 ms^(-2))`

Options

A. `100 m`
B. `90 m`
C. `80 m`
D. `90 m`

Answer

To solve the problem step by step, we will analyze the motion of both stones and use the equations of motion.

### Step 1: Determine the time taken for the first stone to reach the ground.
The height of the hill is given as \( h = 180 \, \text{m} \). The first stone is dropped from this height. We can use the equation of motion:

\[ S = ut + \frac{1}{2} g t^2 \]

Since the stone is dropped (initial velocity \( u = 0 \)), the equation simplifies to:

\[ S = \frac{1}{2} g t^2 \]

Substituting the values:

\[ 180 = \frac{1}{2} \times 10 \times t^2 \]

This simplifies to:

\[ 180 = 5 t^2 \]

Dividing both sides by 5:

\[ t^2 = \frac{180}{5} = 36 \]

Taking the square root:

\[ t = 6 \, \text{s} \]

### Step 2: Determine the time taken for the second stone to reach the ground.
The second stone is dropped 2 seconds after the first stone. Therefore, the time taken by the second stone to reach the ground is:

\[ t_2 = t - 2 = 6 - 2 = 4 \, \text{s} \]

### Step 3: Calculate the height from which the second stone is dropped.
Let \( h_P \) be the height of point P from the ground. The second stone falls for 4 seconds. Using the same equation of motion:

\[ S = ut + \frac{1}{2} g t^2 \]

Again, since the initial velocity \( u = 0 \):

\[ S = \frac{1}{2} g t^2 \]

Substituting the values for the second stone:

\[ S = \frac{1}{2} \times 10 \times (4)^2 \]

Calculating this gives:

\[ S = \frac{1}{2} \times 10 \times 16 = 5 \times 16 = 80 \, \text{m} \]

### Step 4: Relate the height of point P to the height of the hill.
Since the second stone is dropped from point P, the height of point P from the ground is:

\[ h_P = h - S = 180 - 80 = 100 \, \text{m} \]

### Conclusion
The height of point P from the ground is **100 m**.