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 $t = \frac{2u \sin \alpha}{g}$.

Answer 2

This question asks for the formula giving the total time of flight of a projectile launched at angle \( \alpha \) with initial speed \( u \). Let's derive it using the symmetry of projectile motion and check it against each option.

1. \( t = \dfrac{2u\sin\alpha}{g} \): Projectile motion under gravity is symmetric: the time taken to rise from the ground to the peak equals the time taken to fall back down from the peak to the ground. The time to reach maximum height is found by setting the vertical velocity to zero, \( 0 = u\sin\alpha - gt_{rise} \), giving \( t_{rise} = \dfrac{u\sin\alpha}{g} \). Doubling this rise time, since the fall takes the same duration, gives the total time of flight, \( t = 2t_{rise} = \dfrac{2u\sin\alpha}{g} \), matching this option exactly.
2. \( t = \dfrac{2u\cos\alpha}{g} \): This uses the horizontal component of velocity (\( u\cos\alpha \)) instead of the vertical component. Time of flight is governed entirely by the vertical motion under gravity; the horizontal velocity has no bearing on how long the object stays airborne, so this form is inconsistent with the physics.
3. \( t = \dfrac{2u\tan\alpha}{g} \): The tangent function mixes both velocity components together, but the derivation above shows that only the vertical component \( u\sin\alpha \) determines the time to rise and fall, so a tangent-based formula does not correctly represent the vertical motion.
4. \( t = \dfrac{2u}{g\sin\alpha} \): Placing \( \sin\alpha \) in the denominator would mean that as the launch angle approaches 0 degrees (a nearly horizontal launch), the time of flight would blow up toward infinity, contradicting the physical expectation that a flatter launch angle gives a shorter, not longer, time in the air.

The symmetry argument confirms that doubling the rise time, which depends only on the vertical velocity component, gives the correct total time of flight.

Therefore, the correct answer is \( t = \dfrac{2u\sin\alpha}{g} \).