Question

A body is thrown with a velocity of \( 9.8 \) m/s making an angle of \( 30^\circ \) with the horizontal. It will hit the ground after a time:

• A. \( 3.0 \) s
• B. \( 2.0 \) s
• C. \( 1.5 \) s
• D. \( 1.0 \) s

Hint:

The total time of flight for projectile motion is: \[ T = \frac{2 u \sin \theta}{g} \] This formula is useful for determining how long a projectile remains in the air.

Answer 1

The Correct Option is D

Step 1: {Use the time of flight formula}
Time of flight for a projectile is given by: \[ T = \frac{2 u \sin \theta}{g} \] Step 2: {Substituting values}
\[ T = \frac{2 \times 9.8 \times \sin 30^\circ}{9.8} \] Step 3: {Solve for \( T \)}
\[ T = \frac{2 \times 9.8 \times \frac{1}{2}}{9.8} \] \[ T = 1 { sec} \] Thus, the correct answer is (D) 1.0 s.

Answer 2

Given:
Initial velocity \( u = 9.8 \) m/s
Angle of projection \( \theta = 30^\circ \)
We are to find the total time of flight before the body hits the ground.

Step 1: Use time of flight formula
The formula for time of flight for a projectile launched at angle \( \theta \) is:
\[ T = \frac{2u \sin\theta}{g} \] Where:
- \( u \) = initial speed
- \( g = 9.8 \, \text{m/s}^2 \)
- \( \sin 30^\circ = \frac{1}{2} \)

Step 2: Substitute values
\[ T = \frac{2 \times 9.8 \times \frac{1}{2}}{9.8} = \frac{9.8}{9.8} = 1.0 \, \text{second} \]

Final Answer:
The body will hit the ground after 1.0 second.