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: By the symmetry of projectile motion, when the object lands back at the same height it was launched from, its vertical velocity has the same magnitude as at launch but points downward, so \( v_y(\text{landing}) = -u\sin\alpha \).

Step 2: Apply the first equation of motion for the vertical direction, \( v_y = u\sin\alpha - gt \), and substitute the landing condition: \( -u\sin\alpha = u\sin\alpha - gt \).

Step 3: Solve for \( t \): \( gt = u\sin\alpha + u\sin\alpha = 2u\sin\alpha \).
\[ \boxed{t = \dfrac{2u\sin\alpha}{g}} \]