Answer
The correct answer is:
(a) Projectile launched at $60^{\circ}$
Work Step by Step
The correct answer is:
(a) Projectile launched at $60^{\circ}$
Let $v_0$ be the initial velocity.
The time to reach maximum height is $t = \frac{v_0~sin~\theta}{g}$. The total time of flight is $2t = \frac{2~v_0~sin~\theta}{g}$
Since $sin~60^{\circ} \gt sin~30^{\circ}$, the projectile launched at an angle of $60^{\circ}$ has a longer time of flight.
