A stone is thrown horizontally with velocity g `ms^(-1)` from the top of a tower of height g metre. The velocity with which it hits the ground is (in `ms^(-1)`)

Options

A. g
B. 2g
C. `sqrt(3)`g
D. 4g

Answer

To solve the problem of finding the velocity with which a stone hits the ground when thrown horizontally from a height, we can follow these steps:

### Step 1: Identify the Given Information
- The stone is thrown horizontally with a velocity \( u_x = g \, \text{m/s} \).
- The height of the tower is \( h = g \, \text{m} \).
- The acceleration due to gravity is \( g \, \text{m/s}^2 \).

### Step 2: Calculate the Time of Flight
The time of flight \( t \) for an object falling from a height \( h \) under gravity can be calculated using the formula:
\[ t = \sqrt{\frac{2h}{g}} \]
Substituting the values:
\[ t = \sqrt{\frac{2g}{g}} = \sqrt{2} \, \text{s} \]

### Step 3: Determine the Vertical Velocity Just Before Hitting the Ground
The vertical component of the velocity \( v_y \) just before hitting the ground can be calculated using the first equation of motion:
\[ v_y = u_y + gt \]
Since the initial vertical velocity \( u_y = 0 \):
\[ v_y = 0 + g \cdot t = g \cdot \sqrt{2} \, \text{m/s} \]

### Step 4: Calculate the Resultant Velocity
The resultant velocity \( v \) just before hitting the ground can be found using the Pythagorean theorem, since the horizontal and vertical components are perpendicular to each other:
\[ v = \sqrt{(u_x)^2 + (v_y)^2} \]
Substituting the values:
\[ v = \sqrt{(g)^2 + (g \cdot \sqrt{2})^2} \]
\[ v = \sqrt{g^2 + 2g^2} = \sqrt{3g^2} = g\sqrt{3} \, \text{m/s} \]

### Final Answer
The velocity with which the stone hits the ground is:
\[ v = g\sqrt{3} \, \text{m/s} \]

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