A body is projected at an angle of `30^@` with the horizontal and with a speed of `30 ms^-1`. What is the angle with the horizontal after `1.5 s` ? `(g = 10 ms^-2)`.

Options

A. `0^@`
B. `30^@`
C. `60^@`
D. `90^@`

Explanation

To solve the problem, we need to determine the angle of the projectile with respect to the horizontal after 1.5 seconds of flight. Let's break down the solution step by step.

### Step 1: Identify the initial conditions
The body is projected with:
- Initial speed, \( u = 30 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the horizontal and vertical components of the initial velocity
The initial velocity can be resolved into horizontal and vertical components:
- Horizontal component, \( u_x = u \cos(\theta) = 30 \cos(30^\circ) = 30 \times \frac{\sqrt{3}}{2} = 15\sqrt{3} \, \text{m/s} \)
- Vertical component, \( u_y = u \sin(\theta) = 30 \sin(30^\circ) = 30 \times \frac{1}{2} = 15 \, \text{m/s} \)

### Step 3: Calculate the vertical velocity after 1.5 seconds
The vertical velocity at any time \( t \) can be calculated using the formula:
\[ v_y = u_y - g t \]
Substituting the values:
\[ v_y = 15 - 10 \times 1.5 = 15 - 15 = 0 \, \text{m/s} \]

### Step 4: Determine the horizontal velocity
The horizontal velocity remains constant throughout the motion since there is no horizontal acceleration:
\[ v_x = u_x = 15\sqrt{3} \, \text{m/s} \]

### Step 5: Calculate the angle with the horizontal after 1.5 seconds
The angle \( \phi \) with the horizontal can be found using the tangent function:
\[ \tan(\phi) = \frac{v_y}{v_x} \]
Substituting the values we found:
\[ \tan(\phi) = \frac{0}{15\sqrt{3}} = 0 \]
Since \( \tan(\phi) = 0 \), this implies:
\[ \phi = 0^\circ \]

### Final Answer
The angle with the horizontal after 1.5 seconds is \( 0^\circ \).

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