A bullet is fired with a velocity u making an angle of 30° with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is:

Question

A bullet is fired with a velocity u making an angle of 30° with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is: (A) (u/√2) (B) u (C) u√3 (D) (√3/2)u

Tardigrade Admin · Accepted Answer

Horizontal component of the velocity does not change in over all motion of projectile. So, at highest point it will be. u cos θ = u cos 30° =(u√3/2)

Q. A bullet is fired with a velocity u making an angle of $3 0^{\circ}$ with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is:

$\frac{u}{2}$

$u$

$u 3$

$\frac{3}{2} u$

Horizontal component of the velocity does not change in over all motion of projectile.
So, at highest point it will be.
$u$ cos $θ = u$ cos $3 0^{\circ}$
$= \frac{u 3}{2}$

Q. A bullet is fired with a velocity u making an angle of 30∘ with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is:

2​u​

u

u3​

23​​u

Horizontal component of the velocity does not change in over all motion of projectile. So, at highest point it will be. u cos θ=u cos 30∘ =2u3​​

Q. A bullet is fired with a velocity u making an angle of $3 0^{\circ}$ with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is:

$\frac{u}{2}$

$u$

$u 3$

$\frac{3}{2} u$

Horizontal component of the velocity does not change in over all motion of projectile.
So, at highest point it will be.
$u$ cos $θ = u$ cos $3 0^{\circ}$
$= \frac{u 3}{2}$