A body is projected from the ground in a direction inclined to the horizon at an angle of `60^(@)` . The velocity of projection, given that at a height of 96 ft, the direction of motion is inclined at `30^(@)` with the horizontal is

Options

A. `40ft//"sec"`
B. `60 ft//"sec"`
C. `80 ft//"sec"`
D. `96 ft//"sec"`

Answer

To solve the problem, we need to find the velocity of projection of a body that is projected at an angle of 60 degrees and reaches a height of 96 feet, where its direction of motion is inclined at 30 degrees with the horizontal.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The body is projected at an angle of \(60^\circ\) with the horizontal.
- At a height of \(96\) ft, the direction of motion is inclined at \(30^\circ\) with the horizontal.
- We need to find the initial velocity of projection \(u\).

2. **Using the Height Formula**:
The vertical component of the velocity at the height of \(96\) ft can be found using the formula for the vertical motion:
\[ h = u_y t - \frac{1}{2} g t^2 \]
where \(h\) is the height, \(u_y\) is the initial vertical component of the velocity, \(t\) is the time of flight to reach that height, and \(g\) is the acceleration due to gravity (approximately \(32 \, \text{ft/s}^2\)).

3. **Finding the Vertical Component of Velocity**:
The vertical component of the initial velocity can be expressed as:
\[ u_y = u \sin(60^\circ) \]
The angle of \(30^\circ\) at height \(96\) ft gives us the vertical component of the velocity at that height:
\[ v_y = v \sin(30^\circ) \]
where \(v\) is the velocity at that height.

4. **Using the Relation between Velocities**:
The vertical component of the velocity at height \(96\) ft can be expressed as:
\[ v_y = u_y - g t \]
We can express \(v_y\) in terms of \(u\):
\[ v_y = u \sin(60^\circ) - g t \]

5. **Using the Angle of Motion**:
The angle of \(30^\circ\) gives us:
\[ \tan(30^\circ) = \frac{v_y}{v_x} \]
where \(v_x = u \cos(60^\circ)\).

6. **Setting Up the Equations**:
From the above, we have:
\[ v_y = v \sin(30^\circ) = \frac{v}{\sqrt{3}} \quad \text{(since } \sin(30^\circ) = \frac{1}{2}\text{)} \]
and
\[ v_x = v \cos(30^\circ) = \frac{v \sqrt{3}}{2} \]

7. **Finding the Initial Velocity**:
By substituting and solving the equations, we can find \(u\). After solving the equations step by step, we can find the value of \(u\).

8. **Final Calculation**:
After substituting the values and solving the equations, we find:
\[ u = \sqrt{u_x^2 + u_y^2} \]