In a projectile motion, velocity at maximum height is

The correct answer is u cosθ.

To understand why the velocity of a projectile at its maximum height is $ucos\theta$, let's break down the motion of the projectile into horizontal and vertical components.

Suppose a projectile is launched with an initial velocity $u$ at an angle $\theta$ with the horizontal. We can resolve the initial velocity into two perpendicular components:
1. Horizontal component of velocity: $u_x=ucos\theta$
2. Vertical component of velocity: $u_y=usin\theta$

In ideal projectile motion (ignoring air resistance), there is no acceleration acting in the horizontal direction ($a_x=0$). Therefore, the horizontal component of velocity remains constant throughout the flight:
$v_x=u_x=ucos\theta$

In the vertical direction, gravity acts downwards with a constant acceleration ($a_y=-g$). This causes the vertical velocity to decrease as the projectile rises.

When the projectile reaches its maximum height, its upward motion stops momentarily before it starts descending. Consequently, the vertical component of velocity at the highest point becomes zero:
$v_y=0$

The net velocity $v$ of the projectile at any point is the vector sum of its horizontal and vertical components:
$v=\sqrt{{v_x}^2+{v_y}^2}$

Substituting the values at maximum height:
$v=\sqrt{{\left(ucos\theta \right)}^2+0^2}$
$v=ucos\theta$

Thus, at the maximum height, the projectile has only horizontal velocity, which is equal to $ucos\theta$.