A ball is projected with a velocity of $10 ext{ m/s}$ at an angle of $60^{\circ}$ with the vertical directio

Question

A ball is projected with a velocity of $10 	ext{ m/s}$ at an angle of $60^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be:

Accepted Answer

1.  **Analyze the Geometry:** The ball is projected at an angle of $60^{\circ}$ with the *vertical*. Therefore, the angle of projection $	heta$ with the *horizontal* is:
    $$ 	heta = 90^{\circ} - 60^{\circ} = 30^{\circ} $$
2.  **Analyze Velocity at Highest Point:** In projectile motion, the horizontal component of velocity ($u_x$) remains constant throughout the flight because there is no acceleration in the horizontal direction (neglecting air resistance). At the highest point of the trajectory, the vertical component of velocity ($v_y$) becomes zero. Thus, the total speed at the highest point is equal to the horizontal component of the initial velocity.
3.  **Calculate Speed:**
    $$ v_{	ext{highest}} = u_x = u \cos 	heta $$
    Given $u = 10 	ext{ m/s}$ and $	heta = 30^{\circ}$:
    $$ v_{	ext{highest}} = 10 \cos(30^{\circ}) $$
    $$ v_{	ext{highest}} = 10 	imes \frac{\sqrt{3}}{2} = 5\sqrt{3} 	ext{ m/s} $$

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