A body is projected horizontally from the top of a tower with a velocity of 30 m/s. The velocity of the body 4 seconds after projection is (g = `10ms^(-2)`)

Options

A. `40 ms^(-1)`
B. `20 ms^(-1)`
C. `50 ms^(-1)`
D. `100 ms^(-1)`

Answer

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

### Step 1: Identify the initial conditions
The body is projected horizontally with an initial velocity of \( u = 30 \, \text{m/s} \). Since it is projected horizontally, the initial vertical velocity \( u_y = 0 \, \text{m/s} \).

### Step 2: Analyze horizontal motion
In horizontal motion, there is no acceleration (assuming air resistance is negligible). Therefore, the horizontal component of the velocity remains constant:
\[ v_x = u_x = 30 \, \text{m/s} \]

### Step 3: Analyze vertical motion
In vertical motion, the body is subject to gravitational acceleration \( g = 10 \, \text{m/s}^2 \). The vertical velocity after time \( t \) can be calculated using the equation:
\[ v_y = u_y + g \cdot t \]
Substituting the values:
\[ v_y = 0 + 10 \cdot 4 = 40 \, \text{m/s} \]

### Step 4: Combine horizontal and vertical components
The total velocity vector \( \vec{v} \) after 4 seconds can be expressed as:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} = 30 \hat{i} + 40 \hat{j} \, \text{m/s} \]

### Step 5: Calculate the magnitude of the velocity
To find the magnitude of the resultant velocity, we use the Pythagorean theorem:
\[ |\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \text{m/s} \]

### Final Answer
The velocity of the body 4 seconds after projection is \( 50 \, \text{m/s} \).

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