Two second after projection, a projectile is moving at `30^(@)` above the hoorizontal, after one more second it is moving horizontally. The intialy speed of the projectile is `(g = 10 m//s ^(2))`

Options

A. 10 m/s
B. 34 m/s
C. 30 m/s
D. 40 m/s

Answer

To solve the problem step by step, we will analyze the motion of the projectile at different times and use the information given to find the initial speed of the projectile.

### Step 1: Understanding the Motion
We know that:
- At \( t = 2 \) seconds, the projectile is moving at an angle of \( 30^\circ \) above the horizontal.
- At \( t = 3 \) seconds, the projectile is moving horizontally, which means the vertical component of its velocity is zero.

### Step 2: Components of Velocity
Let the initial speed of the projectile be \( u \).
- The initial velocity can be broken down into horizontal and vertical components:
\[ u_x = u \cos \theta \]
\[ u_y = u \sin \theta \]

### Step 3: Velocity at \( t = 2 \) seconds
At \( t = 2 \) seconds, the velocity components are:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y - g \cdot t \)
\[ v_y = u \sin \theta - g \cdot 2 \]

Since the angle is \( 30^\circ \):
\[ v_y = u \sin 30^\circ - 20 = \frac{u}{2} - 20 \]

### Step 4: Setting Up the Equation
At \( t = 2 \) seconds, the angle of the velocity is \( 30^\circ \):
\[ \tan(30^\circ) = \frac{v_y}{u_x} \]
\[ \frac{1}{\sqrt{3}} = \frac{\frac{u}{2} - 20}{u \cos 30^\circ} \]
Using \( \cos 30^\circ = \frac{\sqrt{3}}{2} \):
\[ \frac{1}{\sqrt{3}} = \frac{\frac{u}{2} - 20}{u \cdot \frac{\sqrt{3}}{2}} \]
Cross-multiplying gives:
\[ \frac{u}{2} - 20 = \frac{u}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \]
This simplifies to:
\[ \frac{u}{2} - 20 = \frac{u}{2} \]
We can rearrange this to find \( u \).

### Step 5: Finding the Vertical Component at \( t = 3 \) seconds
At \( t = 3 \) seconds, the vertical component of the velocity is zero:
\[ 0 = u \sin \theta - g \cdot 3 \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ 0 = u \sin \theta - 30 \]
Thus, we have:
\[ u \sin \theta = 30 \]

### Step 6: Using the Horizontal Component
From the previous steps, we have:
1. \( u \sin 30^\circ = 30 \)
2. \( u \cos 30^\circ = u \cdot \frac{\sqrt{3}}{2} \)

### Step 7: Solving for \( u \)
Using the equations:
1. \( u \sin \theta = 30 \)
2. \( u \cos \theta = 10\sqrt{3} \)

Now squaring both equations:
\[ (u \sin \theta)^2 + (u \cos \theta)^2 = 30^2 + (10\sqrt{3})^2 \]
\[ u^2 (\sin^2 \theta + \cos^2 \theta) = 900 + 300 \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ u^2 = 1200 \]
\[ u = \sqrt{1200} = 10\sqrt{12} \approx 34.64 \, \text{m/s} \]

### Final Answer
Thus, the initial speed of the projectile is approximately \( 34 \, \text{m/s} \).

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