Use the figure to answer the question. The figure represents the parabolic trajectory of a ball going from a to e in Earth gravity but without air resistance. The initial velocity of the ball is V o at an angle to the horizon. The vertical dashed lines represent equal time interval, -The vertical velocity at point b is: Note: (g = 9.81 m/s²) A)0 B)V_o cos ($\theta$)- ½ g (2$\Delta$t)² C)V_o sin ($\theta$)- ½ g (2$\Delta$t)² D)V_o sin ($\theta$)+ ½ g (2$\Delta$t)² E)V_o cos ($\theta$)+ ½ g (2$\Delta$t)²

Question

Quizplus · Accepted Answer

The vertical velocity at any point on a parabolic trajectory is equal to the initial vertical velocity minus the acceleration due to gravity multiplied by the time elapsed. At point b, the time elapsed is equal to half the total flight time (indicated by the vertical dashed lines being evenly spaced), which can be determined by finding the horizontal distance travelled from a to b and dividing it by the initial horizontal velocity. Therefore, the vertical velocity at point b is equal to V_o sin (11eca054_3c8a_5b29_a55c_4b66cd642b23_TB6079_11) - 1/2 g (2F1F1F1S1ΔF1F1F10t) = V_o sin (11eca054_3c8a_5b29_a55c_4b66cd642b23_TB6079_11) - g (F1F1F1S1ΔF1F1F10t) = V_o sin (11eca054_3c8a_5b29_a55c_4b66cd642b23_TB6079_11) - 1/2 g (F1F1F1S1ΔF1F1F10/F1F1F1_o) = V_o sin (11eca054_3c8a_5b29_a55c_4b66cd642b23_TB6079_11) - 1/2 g (F1F1F1S1G1F1F1F10/F1F1F1_o) = V_o sin (11eca054_3c8a_5b29_a55c_4b66cd642b23_TB6079_11) - 1/2 g (2F1F1F1S1H1F1F1F10/F1F1F1_o), which is closest to choice C.

Use the Figure to Answer the Question θ\thetaθ )- ½ G (2

Use the Figure to Answer the Question $\theta$ )- ½ G (2