A body is projected horizontally from a very high tower with speed `20 ms ^(-1)` The approximate displacement of the body after 5 s is :

Options

A. 80 m
B. 120 m
C. 160 m
D. 320 m

Explanation

To solve the problem of a body projected horizontally from a high tower with an initial speed of 20 m/s, we can follow these steps:

### Step 1: Understand the Motion
The body is projected horizontally, which means its initial vertical velocity is 0 m/s. The only forces acting on it after projection are gravity, which will affect its vertical motion.

### Step 2: Define the Initial Velocity
The initial velocity vector can be expressed as:
- \( \vec{u} = 20 \hat{i} + 0 \hat{j} \)
Where \( \hat{i} \) represents the horizontal direction and \( \hat{j} \) represents the vertical direction.

### Step 3: Determine the Acceleration
The only acceleration acting on the body is due to gravity, which can be represented as:
- \( \vec{a} = 0 \hat{i} - 10 \hat{j} \)
Here, we take \( g = 10 \, \text{m/s}^2 \) and the negative sign indicates that it acts downward.

### Step 4: Calculate the Displacement
Using the equation of motion for displacement:
\[ \vec{s} = \vec{u} t + \frac{1}{2} \vec{a} t^2 \]
Substituting the values:
- \( t = 5 \, \text{s} \)
- \( \vec{u} = 20 \hat{i} + 0 \hat{j} \)
- \( \vec{a} = 0 \hat{i} - 10 \hat{j} \)

Calculating the displacement:
\[ \vec{s} = (20 \hat{i} + 0 \hat{j}) \cdot 5 + \frac{1}{2} (0 \hat{i} - 10 \hat{j}) \cdot (5^2) \]
\[ \vec{s} = 100 \hat{i} + 0 \hat{j} - \frac{1}{2} (0 \hat{i} - 10 \hat{j}) \cdot 25 \]
\[ \vec{s} = 100 \hat{i} - 125 \hat{j} \]

### Step 5: Calculate the Magnitude of the Displacement
To find the magnitude of the displacement vector \( \vec{s} = 100 \hat{i} - 125 \hat{j} \):
\[ |\vec{s}| = \sqrt{(100)^2 + (-125)^2} \]
\[ |\vec{s}| = \sqrt{10000 + 15625} \]
\[ |\vec{s}| = \sqrt{25625} \]
\[ |\vec{s}| \approx 160 \, \text{m} \]

### Final Answer
The approximate displacement of the body after 5 seconds is **160 meters**.

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