At t = 0,a particle leaves the origin with a velocity of 5.0 m/s in the positive y direction.Its acceleration is given by $\overrightarrow { \mathbf { a } }$ = (3.0 i? - 2.0j?)m/s².At the instant the particle reaches its maximum y coordinate how far is the particle from the origin? A)11 m B)16 m C)22 m D)29 m E)19 m

Question

Quizplus · Accepted Answer

Explanation:At the maximum y coordinate, the particle's velocity in the y direction is zero. So,$$v_{yf}^2=v_{yi}^2+2a_y(y_f-y_i)$$$$0=(5.0	ext{ m/s})^2+2(-2.0	ext{ m/s}^2)(y_f-0)$$$$y_f=6.25	ext{ m}$$The distance from the origin is:$$r=\sqrt{x_f^2+y_f^2}=\sqrt{(3.0	ext{ m/s}^2)(6.25	ext{ s})^2+(6.25	ext{ m})^2}=\boxed{11	ext{ m}}$$

At T = 0,a Particle Leaves the Origin with a Velocity