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 of finding the speed of a body projected horizontally with an initial speed of \(20 \, \text{m/s}\) after \(5 \, \text{s}\), we can follow these steps:

### Step 1: Understand the motion
The body is projected horizontally, which means its initial vertical velocity (\(v_{y0}\)) is \(0 \, \text{m/s}\). The horizontal velocity (\(v_{x}\)) remains constant because there is no horizontal acceleration.

### Step 2: Determine the horizontal velocity
Since there is no horizontal acceleration, the horizontal velocity after \(5 \, \text{s}\) is the same as the initial horizontal velocity:
\[ v_{x} = 20 \, \text{m/s} \]

### Step 3: Calculate the vertical velocity
The vertical motion is influenced by gravity. The vertical velocity (\(v_{y}\)) after \(5 \, \text{s}\) can be calculated using the first equation of motion:
\[ v_{y} = v_{y0} + a \cdot t \]
Here, \(v_{y0} = 0 \, \text{m/s}\) (initial vertical velocity), \(a = -g = -9.8 \, \text{m/s}^2\) (acceleration due to gravity), and \(t = 5 \, \text{s}\).

Substituting the values:
\[ v_{y} = 0 + (-9.8) \cdot 5 = -49 \, \text{m/s} \]
(Note: The negative sign indicates the direction is downward.)

### Step 4: Calculate the resultant speed
The resultant speed (\(v\)) of the body after \(5 \, \text{s}\) can be found using the Pythagorean theorem, since the horizontal and vertical motions are perpendicular to each other:
\[ v = \sqrt{v_{x}^2 + v_{y}^2} \]
Substituting the values:
\[ v = \sqrt{(20)^2 + (-49)^2} \]
\[ v = \sqrt{400 + 2401} = \sqrt{2801} \]

### Step 5: Approximate the resultant speed
Calculating the square root:
\[ v \approx 52.9 \, \text{m/s} \]

### Conclusion
The speed of the body after \(5 \, \text{s}\) is approximately \(52.9 \, \text{m/s}\).

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