A body is projected down from height of 60 m with a velocity `10 ms ^(-1)` at angle `30^(@)` to horizontal. The time of flight of the body is `[g = 10 ms ^(-2)]`

Options

A. 1s
B. 2s
C. 3s
D. 4s

Answer

To solve the problem of a body projected downward from a height of 60 m with an initial velocity of 10 m/s at an angle of 30 degrees to the horizontal, we will follow these steps:

### Step 1: Resolve the initial velocity into components
The initial velocity (u) is given as 10 m/s at an angle of 30 degrees. We need to resolve this into horizontal (u_x) and vertical (u_y) components.

- \( u_x = u \cdot \cos(30^\circ) \)
- \( u_y = u \cdot \sin(30^\circ) \)

Calculating these components:
- \( u_x = 10 \cdot \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \)
- \( u_y = 10 \cdot \sin(30^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \)

### Step 2: Use the equation of motion to find time of flight
We will use the second equation of motion in the vertical direction to find the time of flight (T). The equation is:

\[ h = u_y \cdot T + \frac{1}{2} g T^2 \]

Where:
- \( h = 60 \, \text{m} \) (the height from which the body is projected)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( u_y = 5 \, \text{m/s} \)

Substituting the values into the equation:

\[ 60 = 5T + \frac{1}{2} \cdot 10 \cdot T^2 \]

This simplifies to:

\[ 60 = 5T + 5T^2 \]

Rearranging gives us:

\[ 5T^2 + 5T - 60 = 0 \]

Dividing the entire equation by 5:

\[ T^2 + T - 12 = 0 \]

### Step 3: Solve the quadratic equation
Now we will solve the quadratic equation \( T^2 + T - 12 = 0 \) using the quadratic formula:

\[ T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Where \( a = 1, b = 1, c = -12 \):

Calculating the discriminant:

\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-12) = 1 + 48 = 49 \]

Now substituting back into the quadratic formula:

\[ T = \frac{-1 \pm \sqrt{49}}{2 \cdot 1} \]
\[ T = \frac{-1 \pm 7}{2} \]

This gives us two potential solutions:

1. \( T = \frac{6}{2} = 3 \, \text{s} \)
2. \( T = \frac{-8}{2} = -4 \, \text{s} \) (not valid since time cannot be negative)

### Final Answer
Thus, the time of flight of the body is **3 seconds**.

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