From the top of a tower, a body is thrown horizontally with a speed of `100 ms^(-1)` after 6 s what will be its
(a) Position ? (b) velocity?

Answer

To solve the problem step by step, we will analyze the motion of the body thrown horizontally from the top of a tower.

### Given Data:
- Horizontal speed (u_x) = 100 m/s
- Time (t) = 6 s
- Acceleration due to gravity (g) = 10 m/s² (approximately)

### Step 1: Calculate the horizontal position (x)
Since the body is thrown horizontally, its horizontal velocity remains constant as there is no horizontal acceleration.

**Formula:**
\[ x = u_x \cdot t \]

**Calculation:**
\[ x = 100 \, \text{m/s} \cdot 6 \, \text{s} = 600 \, \text{m} \]

### Step 2: Calculate the vertical position (y)
The vertical position can be calculated using the formula for displacement under constant acceleration.

**Formula:**
\[ y = u_y \cdot t + \frac{1}{2} g t^2 \]
Where:
- \( u_y \) (initial vertical velocity) = 0 (since it is thrown horizontally)

**Calculation:**
\[ y = 0 + \frac{1}{2} \cdot 10 \, \text{m/s}^2 \cdot (6 \, \text{s})^2 \]
\[ y = \frac{1}{2} \cdot 10 \cdot 36 = 180 \, \text{m} \]

### Step 3: Calculate the vertical velocity (v_y) after 6 seconds
The vertical velocity can be calculated using the formula:

**Formula:**
\[ v_y = u_y + g \cdot t \]

**Calculation:**
\[ v_y = 0 + 10 \, \text{m/s}^2 \cdot 6 \, \text{s} = 60 \, \text{m/s} \]

### Step 4: Calculate the resultant velocity (v)
The resultant velocity can be calculated using the Pythagorean theorem, as the horizontal and vertical components are perpendicular to each other.

**Formula:**
\[ v = \sqrt{u_x^2 + v_y^2} \]

**Calculation:**
\[ v = \sqrt{(100 \, \text{m/s})^2 + (60 \, \text{m/s})^2} \]
\[ v = \sqrt{10000 + 3600} \]
\[ v = \sqrt{13600} \]
\[ v \approx 116.62 \, \text{m/s} \]

### Final Answers:
(a) Position after 6 seconds:
- Horizontal position (x) = 600 m
- Vertical position (y) = 180 m

(b) Velocity after 6 seconds:
- Resultant velocity (v) ≈ 116.62 m/s

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