A body of mass 1kg is projected with velocity `50 ms^(-1)` at an angle of `30^(@)` with the horizontal. At the highest point of its path a force 10 N starts acting on body for 5s vertically upward besids gravitational force, what is horizontal range of the body (Take, `g = 10 ms^(-2))`

Options

A. `125 sqrt(3)`
B. `200 sqrt(3)m`
C. `500m`
D. `250 sqrt(3)m`

Explanation

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

### Step 1: Determine the initial horizontal and vertical components of the velocity
The initial velocity \( u \) is given as \( 50 \, \text{m/s} \) and the angle \( \theta \) is \( 30^\circ \).

- The horizontal component of the velocity \( u_x \) is given by:
\[ u_x = u \cos \theta = 50 \cos(30^\circ) = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \, \text{m/s} \]

- The vertical component of the velocity \( u_y \) is given by:
\[ u_y = u \sin \theta = 50 \sin(30^\circ) = 50 \times \frac{1}{2} = 25 \, \text{m/s} \]

### Step 2: Calculate the time to reach the highest point
At the highest point, the vertical velocity becomes zero. We can use the formula:
\[ v_y = u_y - g t \]
Setting \( v_y = 0 \):
\[ 0 = 25 - 10t \implies t = \frac{25}{10} = 2.5 \, \text{s} \]

### Step 3: Total time of flight
The total time of flight for a projectile is twice the time taken to reach the highest point:
\[ T = 2 \times 2.5 = 5 \, \text{s} \]

### Step 4: Calculate the distance traveled during the upward force application
After reaching the highest point, a force of \( 10 \, \text{N} \) acts vertically upward for \( 5 \, \text{s} \). The net force acting on the body during this time is:
\[ F_{\text{net}} = F_{\text{up}} - F_{\text{gravity}} = 10 \, \text{N} - 10 \, \text{N} = 0 \, \text{N} \]
Since the net force is zero, the body continues to move horizontally with the horizontal component of the velocity \( u_x \).

The horizontal distance traveled during this time is:
\[ \text{Distance} = u_x \times t = 25\sqrt{3} \times 5 = 125\sqrt{3} \, \text{m} \]

### Step 5: Calculate the range of the projectile
The horizontal range of the projectile can be calculated using the formula for the range of a projectile:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Substituting the values:
\[ R = \frac{50^2 \sin(60^\circ)}{10} = \frac{2500 \times \frac{\sqrt{3}}{2}}{10} = 125\sqrt{3} \, \text{m} \]

### Step 6: Total horizontal range
The total horizontal range is the sum of the range of the projectile and the distance traveled during the upward force application:
\[ \text{Total Range} = R + \text{Distance} = 125\sqrt{3} + 125\sqrt{3} = 250\sqrt{3} \, \text{m} \]

### Final Answer
The horizontal range of the body is:
\[ \text{Total Range} = 250\sqrt{3} \, \text{m} \]

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