Projectile Motion

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A projectile is launched from level ground with a horizontal velocity component of $18\text{ m/s}$ and an initial vertical velocity component of $24\text{ m/s}$ . Neglecting air resistance, what is the magnitude of the velocity of the projectile at the apex of its trajectory?

0\text{ m/s}

18\text{ m/s}

24\text{ m/s}

30\text{ m/s}

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Cevap: $18\text{ m/s}$

Cevap

18\text{ m/s}

At the apex of a projectile's flight, vertical motion momentarily halts (

v_y = 0\text{ m/s}

), but horizontal motion continues unchanged (

v_x = 18\text{ m/s}

). Thus, the magnitude of the velocity at the apex equals the horizontal velocity component,

18\text{ m/s}

Adım Adım Çözüm

Analyze the velocity components at the trajectory apex (highest point)

Vertical velocity component

v_y = 0\text{ m/s}

and horizontal velocity component

v_x = u_x = 18\text{ m/s}

Gravity acts downward, reducing the vertical speed component to zero at maximum height. In the absence of air resistance, no horizontal forces act on the projectile, so

v_x

remains constant throughout the flight.

Calculate the magnitude of the total velocity at the apex

v = \sqrt{v_x^2 + v_y^2} = \sqrt{18^2 + 0^2} = 18\text{ m/s}

The total velocity magnitude is obtained by vector addition of its orthogonal components.

Anahtar Kavram

Independence of horizontal and vertical motion components in projectile motion.

Tahmini Süre:1m 0s

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Projectile Motion

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