A body is thrown with velocity (4i+3j) metre per second. Its horizontal range and time of flight are

Options

A. 2.4 m and 0.6 s
B. 3.2 m and 0.8 s
C. 5 m and 2 s
D. 7 m and 1s

Answer

To solve the problem of finding the horizontal range and time of flight for a body thrown with an initial velocity of \( \vec{u} = 4\hat{i} + 3\hat{j} \) m/s, we can follow these steps:

### Step 1: Determine the Magnitude of the Initial Velocity
The magnitude of the initial velocity \( u \) can be calculated using the formula:
\[ u = \sqrt{u_x^2 + u_y^2} \]
where \( u_x = 4 \) m/s (horizontal component) and \( u_y = 3 \) m/s (vertical component).

Calculating the magnitude:
\[ u = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ m/s} \]

### Step 2: Calculate the Angle of Projection
The angle \( \theta \) of projection can be found using the tangent function:
\[ \tan(\theta) = \frac{u_y}{u_x} = \frac{3}{4} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.86^\circ \]

### Step 3: Calculate the Horizontal Range
The formula for the horizontal range \( R \) in projectile motion is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

Calculating \( \sin(2\theta) \):
\[ \sin(2\theta) = \sin(2 \times 36.86^\circ) \approx \sin(73.72^\circ) \approx 0.956 \]

Now substituting the values into the range formula:
\[ R = \frac{5^2 \times 0.956}{10} = \frac{25 \times 0.956}{10} = \frac{23.9}{10} \approx 2.39 \text{ m} \]

### Step 4: Calculate the Time of Flight
The time of flight \( T \) can be calculated using the formula:
\[ T = \frac{2u_y}{g} \]
Substituting the values:
\[ T = \frac{2 \times 3}{10} = \frac{6}{10} = 0.6 \text{ seconds} \]

### Final Answers
- Horizontal Range \( R \approx 2.39 \, \text{m} \)
- Time of Flight \( T = 0.6 \, \text{s} \)