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 can follow these steps:

### Step 1: Resolve the Initial Velocity
The initial velocity \( u \) is given as 10 m/s at an angle of 30 degrees. We need to find the vertical component of this velocity since the motion in the vertical direction is what will determine the time of flight.

- The vertical component \( u_y \) can be calculated using:
\[ u_y = u \sin(\theta) = 10 \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s} \]

### Step 2: Set Up the Equation of Motion
We will use the second equation of motion for vertical motion, which is:
\[ h = u_y t + \frac{1}{2} a t^2 \]
Where:
- \( h = 60 \, \text{m} \) (the height from which the body is projected)
- \( u_y = 5 \, \text{m/s} \) (the initial vertical velocity)
- \( a = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity, acting downward)
- \( t \) is the time of flight we need to find.

### Step 3: Substitute Values into the Equation
Substituting the known 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 4: Solve the Quadratic Equation
Now we can solve the quadratic equation \( t^2 + t - 12 = 0 \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = 1, c = -12 \):
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \]
\[ t = \frac{-1 \pm \sqrt{1 + 48}}{2} \]
\[ t = \frac{-1 \pm \sqrt{49}}{2} \]
\[ t = \frac{-1 \pm 7}{2} \]
This gives us two possible solutions for \( t \):
\[ t = \frac{6}{2} = 3 \quad \text{and} \quad t = \frac{-8}{2} = -4 \]

### Step 5: Determine the Valid Time
Since time cannot be negative, we discard \( t = -4 \) and accept:
\[ t = 3 \, \text{seconds} \]

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

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