Question

The time of flight (t) of a projectile on a horizontal plane is given by: ____.

• A. t = 2u sin $\alpha$ g
• B. t = 2u cos $\alpha$ g
• C. t = 2u tan $\alpha$ g
• D. t = 2u (g sin $\alpha$)

Hint:

The time taken to reach the maximum height is exactly half of the total time of flight, which is $u \sin \alpha / g$. What goes up must come down in the same amount of time in an ideal vacuum!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The time of flight is the total time the projectile remains in the air before hitting the ground. It is determined by the vertical component of the initial velocity and the acceleration due to gravity.

Step 2: Key Formula or Approach:
The vertical component of initial velocity is $u_y = u \sin \alpha$. The motion ends when the vertical displacement ($y$) becomes zero.

Step 3: Detailed Explanation:
Using the second equation of motion for the vertical direction: \[ y = u_yt - \frac{1}{2}gt^2 \] Set $y = 0$ for the full flight: \[ 0 = (u \sin \alpha)t - \frac{1}{2}gt^2 \] \[ \frac{1}{2}gt^2 = (u \sin \alpha)t \] Dividing by $t$ (since $t \neq 0$): \[ \frac{1}{2}gt = u \sin \alpha \] \[ t = \frac{2u \sin \alpha}{g} \]

Step 4: Final Answer:
The time of flight is given by $t = \frac{2u \sin \alpha}{g}$.