A ball is projected from point $A$ with velocity $20 ext{ m s}^{-1}$ at an angle $60^{\circ}$ to the horizon

Question

A ball is projected from point $A$ with velocity $20 	ext{ m s}^{-1}$ at an angle $60^{\circ}$ to the horizontal direction. At the highest point $B$ of the path (as shown in figure), the velocity $v$ (in $	ext{m s}^{-1}$) of the ball will be:

Accepted Answer

1.  **Analyze Velocity Components:** In projectile motion, the velocity vector can be resolved into horizontal ($u_x$) and vertical ($u_y$) components. The horizontal acceleration is zero ($a_x = 0$), so the horizontal component of velocity remains constant throughout the flight.
2.  **Velocity at Highest Point:** At the highest point of the trajectory, the vertical component of the velocity becomes zero ($v_y = 0$). Therefore, the total velocity at the highest point is entirely horizontal.
    $$ v_{	ext{top}} = u_x = u \cos 	heta $$
3.  **Calculation:**
    Given initial velocity $u = 20 	ext{ m s}^{-1}$ and angle $	heta = 60^{\circ}$.
    $$ v_{	ext{top}} = 20 \cos(60^{\circ}) $$
    $$ v_{	ext{top}} = 20 	imes 0.5 = 10 	ext{ m s}^{-1} $$

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