Motion In Two Dimensions Question 5
Question 5 - 31 January - Shift 1
The initial speed of a projectile fired from ground is $u$. At the highest point during its motion, the speed of projectile is $\frac{\sqrt{3}}{2} u$. The time of flight of the projectile is:
(1) $\frac{u}{2 g}$
(2) $\frac{u}{g}$
(3) $\frac{2 u}{g}$
(4) $\frac{\sqrt{3} u}{g}$
Show Answer
Answer: (2)
Solution:
Formula: Time of flight
$u \cos \theta=\frac{\sqrt{3} u}{2} \Rightarrow \cos \theta=\frac{\sqrt{3}}{2}$
$\Rightarrow \theta=30^{\circ}$
$T=\frac{2 u \sin 30^{\circ}}{g}=\frac{u}{g}$
