AT&T; LTE 4:05 PM webassign.net initial velocity, upward and to the right, with

Question

AT&T; LTE 4:05 PM webassign.net initial velocity, upward and to the right, with forces from air resistance neglected. There is no force acting horizontally to accelerate its horizontal motion, while its vertical motion is accelerated downward by gravity. Therefore as the projectile moves to the right at a constant rate, the vertical part of its motion consists of first rising upward and then moving downward until the projectile strikes the ground. Use the simulation to display the projectile motion. Categorize The velocity has components in both the x and y directions, so we categorize this as a problem involving particle motion in two dimensions. The particle also has only a y component of acceleration, so we categorize it as a particle under constant acceleration in the y direction and constant velocity in the direction. (A) How do the vertical and horizontal components of its velocity vary with time? Analyze The initial velocity in the r-direction v, is related to the initial speed by Yu=v,cos 50° The constant velocity in the r-direction means that the equation describing the time dependence of x for the particle, with taken as 0, is The equation for the vertical coordinate, which is constantly accelerating downward at g 9.8 m/s, is x m/s) r The correct answer is not zero. Finalize The-gr term is negative. The other time-dependent term is proportional to r and positive. Which of the two dominates at small? Which term's magnitude gets larger faster asr gets large? What effect does that have on the sign of the y coordinate as starts out small and then gets larger? Is this consistent with the path you expect the projectile to take? (B) How long does it remain in flight? The y-component of the projectile's velocity decreases by 9.8

Explanation / Answer

vi Cos50 = 20.52

hence vi = 32 m/s

initial velocity along Y-direction is given as

Voy = vi Sin50 = (32) Sin50 = 24.5 m/s

so the equation for vertical coordinate becomes

y = 24.5 t - 4.9 t2

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