Question

which of these is the correct formula for the time of flight of an angular projectile? a (t=\frac{usin\theta}{g}) b (t = \frac{usin\theta}{2g}) c (t=\frac{2usin\theta}{g})

Explanation:

Step1: Recall projectile - motion formula

The time - of - flight formula for an angular projectile (projectile launched at an angle $\theta$ with initial velocity $u$) is derived from the vertical - motion equation. In vertical motion, the displacement $y = 0$ (when the projectile returns to the same height), and the initial vertical velocity is $u_y=u\sin\theta$. Using the equation $y = u_y t-\frac{1}{2}gt^{2}$, setting $y = 0$, we get $0 = u\sin\theta t-\frac{1}{2}gt^{2}$. Factoring out $t$, we have $t(u\sin\theta-\frac{1}{2}gt)=0$. One solution is $t = 0$ (corresponds to the initial time), and the other is $t=\frac{2u\sin\theta}{g}$.

Answer:

The correct formula for the time of flight of an angular projectile is $T=\frac{2u\sin\theta}{g}$. None of the given options (a. $T = \frac{u\sin\theta}{g}$, b. $T=\frac{u\sin\theta}{2g}$) are correct.