A body is projected at angle 30° to horizontal with a velocity 50 `ms^(-1)` . Its time of flight is (g=10 `m//s^2`)

Options

A. 4s
B. 5s
C. 6s
D. 7s

Answer

To find the time of flight of a body projected at an angle, we can use the formula:

\[ T = \frac{2u \sin(\theta)}{g} \]

where:
- \( T \) is the time of flight,
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

**Step 1: Identify the given values**
- Initial velocity, \( u = 50 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

**Step 2: Calculate \( \sin(\theta) \)**
- We need to find \( \sin(30^\circ) \).
- From trigonometric values, we know that \( \sin(30^\circ) = \frac{1}{2} \).

**Step 3: Substitute the values into the formula**
- Now we can substitute the values into the time of flight formula:
\[ T = \frac{2 \times 50 \times \sin(30^\circ)}{10} \]
\[ T = \frac{2 \times 50 \times \frac{1}{2}}{10} \]

**Step 4: Simplify the expression**
- Simplifying the expression:
\[ T = \frac{2 \times 50 \times 0.5}{10} \]
\[ T = \frac{50}{10} \]
\[ T = 5 \, \text{seconds} \]

**Final Answer:**
The time of flight is \( 5 \, \text{seconds} \).

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