The position of a projectile launched from the origin at t = 0 is given by `s`=`(40hati+50hatj)m`at `t=2s`. if the projectile was launched at an angle `theta` from the horizontal, then `theta`is (take g = 10 `ms^(–2)`

Options

A. (a) ` tan ^( -1) "" (2)/(3) `
B. (b)` tan ^(-1) "" (3)/(2) `
C. (c)` tan ^(-1) "" (7)/(4) `
D. (d)` tan ^(-1) "" (4)/(5) `

Explanation

To solve the problem, we need to find the angle of projection \( \theta \) of a projectile launched from the origin with a given position vector at \( t = 2 \) seconds. The position vector is given as \( \mathbf{s} = 40 \hat{i} + 50 \hat{j} \) meters.

### Step-by-Step Solution:

1. **Identify the Components of Motion**:
The position vector at \( t = 2 \) seconds gives us the horizontal and vertical displacements:
- Horizontal displacement \( s_x = 40 \) m
- Vertical displacement \( s_y = 50 \) m

2. **Calculate Horizontal Velocity**:
The horizontal component of the velocity \( u_x \) can be calculated as:
\[ u_x = \frac{s_x}{t} = \frac{40 \, \text{m}}{2 \, \text{s}} = 20 \, \text{m/s} \]

3. **Use the Vertical Motion Equation**:
For vertical motion, we can use the equation of motion:
\[ s_y = u_y t + \frac{1}{2} a t^2 \]
Here, \( a = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity), and substituting the known values:
\[ 50 = u_y \cdot 2 + \frac{1}{2} (-10) \cdot (2^2) \]
Simplifying the equation:
\[ 50 = 2u_y - 20 \]
\[ 2u_y = 70 \implies u_y = 35 \, \text{m/s} \]

4. **Calculate the Angle of Projection**:
The angle \( \theta \) can be found using the tangent function:
\[ \tan \theta = \frac{u_y}{u_x} = \frac{35}{20} \]
Therefore:
\[ \tan \theta = \frac{7}{4} \]

5. **Find the Angle \( \theta \)**:
To find \( \theta \), we take the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{7}{4}\right) \]

### Final Answer:
Thus, the angle \( \theta \) at which the projectile was launched is:
\[ \theta = \tan^{-1}\left(\frac{7}{4}\right) \]