A body is projected horizontally with speed `30m//s` from a very high tower. What will be its speed after 4sec-

Options

A. `20 m//s`
B. `50 m//s`
C. `54 m//s`
D. `70 m//s`

Explanation

To solve the problem of finding the speed of a body projected horizontally from a high tower after 4 seconds, we can break it down into steps.

### Step-by-Step Solution:

1. **Identify the Components of Motion**:
- The body is projected horizontally with an initial horizontal speed \( u = 30 \, \text{m/s} \).
- There is no initial vertical speed, so the vertical component of velocity at the start is \( v_{y0} = 0 \).

2. **Calculate the Vertical Velocity After 4 Seconds**:
- The vertical motion is influenced by gravity. The vertical velocity after time \( t \) can be calculated using the formula:
\[ v_y = v_{y0} + gt \]
- Where:
- \( g \) is the acceleration due to gravity, approximately \( 10 \, \text{m/s}^2 \).
- \( t = 4 \, \text{s} \).
- Substituting the values:
\[ v_y = 0 + (10 \, \text{m/s}^2)(4 \, \text{s}) = 40 \, \text{m/s} \]

3. **Determine the Resultant Velocity**:
- The resultant velocity \( v \) of the body after 4 seconds can be found using the Pythagorean theorem, since the horizontal and vertical motions are perpendicular to each other.
- The resultant velocity is given by:
\[ v = \sqrt{u^2 + v_y^2} \]
- Substituting the known values:
\[ v = \sqrt{(30 \, \text{m/s})^2 + (40 \, \text{m/s})^2} \]
\[ v = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \text{m/s} \]

4. **Final Answer**:
- The speed of the body after 4 seconds is \( 50 \, \text{m/s} \).