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Difficulty: MediumProjectile Motion

A projectile is launched from level ground with a horizontal velocity component of 18 m/s18\text{ m/s} and an initial vertical velocity component of 24 m/s24\text{ m/s}. Neglecting air resistance, what is the magnitude of the velocity of the projectile at the apex of its trajectory?

  1. A
    0 m/s0\text{ m/s}
  2. 18 m/s18\text{ m/s}Answer
  3. C
    24 m/s24\text{ m/s}
  4. D
    30 m/s30\text{ m/s}

Answer

18 m/s18\text{ m/s}
At the apex of a projectile's flight, vertical motion momentarily halts (vy=0 m/sv_y = 0\text{ m/s}), but horizontal motion continues unchanged (vx=18 m/sv_x = 18\text{ m/s}). Thus, the magnitude of the velocity at the apex equals the horizontal velocity component, 18 m/s18\text{ m/s}.

Step-by-Step Solution

1
Analyze the velocity components at the trajectory apex (highest point)
Vertical velocity component vy=0 m/sv_y = 0\text{ m/s} and horizontal velocity component vx=ux=18 m/sv_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 vxv_x remains constant throughout the flight.
2
Calculate the magnitude of the total velocity at the apex
v=vx2+vy2=182+02=18 m/sv = \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.

Key Concept

Independence of horizontal and vertical motion components in projectile motion.
Estimated Time:1m 0s
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Topics

Projectile Motion

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