A projectile is launched from the edge of a cliff with a velocity of 10.0 m/s at an angle of 30.0 degrees above the horizontal. What is the velocity of the projectile 2.00 seconds later (ignoring air resistance, and assuming it hasn't reached the ground yet)? A) V_x = 8.7 m/s, V_y = −14.6 m/s B) V_x = 14.6 m/s, V_y = −5.6 m/s C) V_x = −8.7 m/s, V_y = −14.6 m/s D) V_x = −14.6 m/s, V_y = 8.7 m/s E) V_x = 5.6 m/s, V_y = −14.6 m/s

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Quizplus · Accepted Answer

The horizontal component of velocity remains constant because there is no air resistance. The vertical component changes due to gravity. Initially, $V_{x0} = V_0 \cos(\theta) = 10 \cos(30^\circ) = 8.66 \, \text{m/s}$ (approximately 8.7 m/s), and $V_{y0} = V_0 \sin(\theta) = 10 \sin(30^\circ) = 5 \, \text{m/s}$. After 2 seconds, the vertical velocity is $V_{y} = V_{y0} - g t = 5 - 9.8 \times 2 = -14.6 \, \text{m/s}$, where $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity. The horizontal velocity remains $8.7 \, \text{m/s}$.

A Projectile Is Launched from the Edge of a Cliff