The initial speed of a projectile fired from ground is u. At the highest point during its motion, the speed of projectile is (√3/2) u. The time of flight of the projectile is :

Question

The initial speed of a projectile fired from ground is u. At the highest point during its motion, the speed of projectile is (√3/2) u. The time of flight of the projectile is : (A) (2 u / g ) (B) ( u /2 g ) (C) ( u / g ) (D) (√3 u/g)

Tardigrade Admin · Accepted Answer

u cos θ=(√3 u /2) ⇒ cos θ=(√3/2) ⇒ θ=30° T =(2 u sin 30°/ g )=( u / g )

Q. The initial speed of a projectile fired from ground is $u$ . At the highest point during its motion, the speed of projectile is $\frac{3}{2} u$ . The time of flight of the projectile is :

$\frac{2 u}{g}$

$\frac{u}{2 g}$

$\frac{u}{g}$

$\frac{3 u}{g}$

$u cos θ = \frac{3 u}{2}$
$\Rightarrow cos θ = \frac{3}{2}$
$\Rightarrow θ = 3 0^{\circ}$
$T = \frac{2 u s i n 3 0 ^{\circ}}{g} = \frac{u}{g}$

Q. The initial speed of a projectile fired from ground is u. At the highest point during its motion, the speed of projectile is 23​​u. The time of flight of the projectile is :

g2u​

2gu​

gu​

g3​u​