If a projectile is launched with 3m/s velocity at 60-degree angle, then at highest point its horizontal velocity is

To determine the horizontal velocity of a projectile at its highest point, we need to consider the components of the initial velocity.

1. Initial velocity components:

   • The initial velocity ($v_0$) is 3 m/s.
   • The angle of projection ($\theta$) is 60 degrees.

2. Horizontal component of the initial velocity:

   • The horizontal component ($v_{0x}$) can be calculated using the cosine function: $v_{0x} = v_0 \cdot \cos(\theta)$
   • Substituting the given values: $v_{0x} = 3 \cdot \cos(60^\circ)$
   • We know that $\cos(60^\circ) = 0.5$: $v_{0x} = 3 \cdot 0.5 = 1.5 \text{ m/s}$

3. Horizontal velocity at the highest point:

   • The horizontal velocity of a projectile remains constant throughout its flight (assuming no air resistance).
   • Therefore, the horizontal velocity at the highest point is the same as the initial horizontal velocity.

Thus, the horizontal velocity at the highest point is $1.5 \text{ m/s}$.