Question

A bullet is fired with a velocity u making an angle of 60° with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is

Answer 1

Your question is incomplete

Answer 2

The answer to the  question is $\frac{u}{2}$ i.e the horizontal component of the velocity when the bullet reaches at the maximum height is $\frac{u}{2}$

CALCULATION:

As per the question, the initial projected velocity of the bullet is u .

The bullet was fired at angle of 60 degree with the horizontal.

Hence, the horizontal component of the initial projected velocity is -

                                $u_{x}\ =\ u.cos\theta$

                                      $=\ u\times cos60$

                                      $=\ \frac{u}{2}$        [cos60 = 0.5]

The bullet is a projectile. Hence, the only force that acts on the bullet is the force of gravity . It acts in vertical downward direction. Hence, the horizontal velocity will not be affected by it. There will be change in the vertical component of the velocity.

As the horizontal component is constant, hence, the horizontal component of the velocity at the maximum height is $\frac{u}{2}$.