A body is thrown from a point with speed `50 ms ^(-1)` at an angle `37^@` with horizontal . When it has moved a horizontal distance of 80 m then its distance from point of projection is

Options

A. 40 m
B. `40 sqrt(2) m`
C. `40 sqrt(5) m`
D. none

Answer

To solve the problem step by step, we will break it down into manageable parts:

### Step 1: Determine the horizontal and vertical components of the initial velocity.
The body is thrown with an initial speed \( v = 50 \, \text{m/s} \) at an angle \( \theta = 37^\circ \).

- **Horizontal Component (\( v_x \))**:
\[ v_x = v \cdot \cos(\theta) = 50 \cdot \cos(37^\circ) \]
Using \( \cos(37^\circ) \approx \frac{4}{5} \):
\[ v_x = 50 \cdot \frac{4}{5} = 40 \, \text{m/s} \]

- **Vertical Component (\( v_y \))**:
\[ v_y = v \cdot \sin(\theta) = 50 \cdot \sin(37^\circ) \]
Using \( \sin(37^\circ) \approx \frac{3}{5} \):
\[ v_y = 50 \cdot \frac{3}{5} = 30 \, \text{m/s} \]

### Step 2: Calculate the time taken to travel a horizontal distance of 80 m.
The horizontal motion is uniform, so the time \( t \) taken to cover the horizontal distance \( x = 80 \, \text{m} \) is given by:
\[ t = \frac{x}{v_x} = \frac{80}{40} = 2 \, \text{s} \]

### Step 3: Calculate the vertical displacement after 2 seconds.
Using the formula for vertical displacement \( s_y \):
\[ s_y = v_y \cdot t + \frac{1}{2}(-g)t^2 \]
where \( g \approx 10 \, \text{m/s}^2 \) (acceleration due to gravity).

Substituting the values:
\[ s_y = 30 \cdot 2 + \frac{1}{2}(-10)(2^2) \]
\[ s_y = 60 - 20 = 40 \, \text{m} \]

### Step 4: Calculate the distance from the point of projection.
Now, we need to find the resultant distance \( d \) from the point of projection using the Pythagorean theorem:
\[ d = \sqrt{x^2 + s_y^2} \]
Substituting \( x = 80 \, \text{m} \) and \( s_y = 40 \, \text{m} \):
\[ d = \sqrt{80^2 + 40^2} = \sqrt{6400 + 1600} = \sqrt{8000} = 40\sqrt{5} \, \text{m} \]

### Final Result
The distance from the point of projection is:
\[ d \approx 89.44 \, \text{m} \]