A stone is projected horizontally with a velocity `9.8 ms ^(-1)` from a tower of height 100 m. Its velocity one second after projection is `(g = 9.8 ms ^(-2)).`

Options

A. `9.8 ms ^(-1)`
B. `4.9 ms ^(-1)`
C. `9.8 sqrt2 ms ^(-1)`
D. `4.9sqrt2 ms ^(-1)`

Answer

To solve the problem step by step, we need to analyze the motion of the stone projected horizontally from a height.

### Step 1: Identify the components of motion
The stone is projected horizontally, which means it has two components of motion:
- Horizontal motion with an initial velocity \( V_x = 9.8 \, \text{m/s} \)
- Vertical motion under the influence of gravity.

### Step 2: Calculate the vertical component of velocity after 1 second
The vertical component of the velocity \( V_y \) after time \( t \) can be calculated using the equation:
\[ V_y = V_{y0} + gt \]
where:
- \( V_{y0} = 0 \) (initial vertical velocity, since it is projected horizontally),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( t = 1 \, \text{s} \).

Substituting the values:
\[ V_y = 0 + (9.8 \, \text{m/s}^2)(1 \, \text{s}) = 9.8 \, \text{m/s} \]

### Step 3: Write the velocity vector after 1 second
The velocity vector \( \vec{V} \) after 1 second can be expressed in terms of its components:
\[ \vec{V} = V_x \hat{i} + V_y \hat{j} \]
Substituting the values we found:
\[ \vec{V} = 9.8 \hat{i} + 9.8 \hat{j} \]

### Step 4: Calculate the magnitude of the velocity vector
The magnitude of the velocity \( V \) can be calculated using the Pythagorean theorem:
\[ V = \sqrt{V_x^2 + V_y^2} \]
Substituting the values:
\[ V = \sqrt{(9.8)^2 + (9.8)^2} = \sqrt{2 \times (9.8)^2} = 9.8 \sqrt{2} \, \text{m/s} \]

### Final Answer
The velocity of the stone one second after projection is:
\[ V = 9.8 \sqrt{2} \, \text{m/s} \]

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