A body is projected horizontally with speed `20 ms^(-1)` . The speed of the body after 5s is nearly

Options

A. `54 ms^(-1)`
B. `35 ms^(-1)`
C. `20 ms^(-1)`
D. `70 ms^(-1)`

Answer

To solve the problem step by step, we will analyze the motion of the body projected horizontally and calculate its speed after 5 seconds.

### Step 1: Understand the motion
The body is projected horizontally with an initial speed of \( V_x = 20 \, \text{m/s} \). Since it is projected horizontally, the initial vertical velocity \( V_{y0} = 0 \, \text{m/s} \).

### Step 2: Determine the vertical motion
The body is subject to gravitational acceleration \( g = 10 \, \text{m/s}^2 \) (assuming standard value for simplicity). We need to find the vertical velocity \( V_y \) after \( t = 5 \, \text{s} \).

Using the first equation of motion:
\[ V_y = V_{y0} + a_y \cdot t \]
Where:
- \( V_{y0} = 0 \, \text{m/s} \) (initial vertical velocity)
- \( a_y = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 5 \, \text{s} \)

Substituting the values:
\[ V_y = 0 + 10 \cdot 5 = 50 \, \text{m/s} \]

### Step 3: Calculate the horizontal velocity
The horizontal velocity \( V_x \) remains constant throughout the motion since there is no horizontal acceleration:
\[ V_x = 20 \, \text{m/s} \]

### Step 4: Calculate the resultant velocity
Now, we need to find the resultant velocity \( V \) after 5 seconds using the Pythagorean theorem:
\[ V = \sqrt{V_x^2 + V_y^2} \]

Substituting the values:
\[ V = \sqrt{(20)^2 + (50)^2} = \sqrt{400 + 2500} = \sqrt{2900} \]

Calculating \( \sqrt{2900} \):
\[ V \approx 54 \, \text{m/s} \]

### Conclusion
The speed of the body after 5 seconds is approximately \( 54 \, \text{m/s} \).

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