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 $\theta$ to the horizon. The vertical dashed lines represent equal time interval, $\Delta$t. Consider "up" as the positive direction. 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

At point b, the ball has reached its highest point in the trajectory, so the vertical velocity is zero. From the conservation of energy, we know that the initial total energy equals the final total energy. At point a, the initial energy is given by: Ea = ½ m V_o² At point b, the potential energy is converted to kinetic energy and we have: Eb = ½ m Vb² where Vb is the velocity magnitude at point b. Therefore, we have: Ea = Eb ½ m V_o² = ½ m Vb² + mgh where h is the height of point b relative to point a. Since h is maximum at point b, we can write: h = V_o² sin² F1F1F1S1F1F1F10) / 2g Substituting h into the previous equation, we get: ½ m V_o² = ½ m Vb² + V_o² sin² F1F1F1S1F1F1F10) / 2 Solving for Vb, we obtain: Vb = V_o sin F1F1F1S1F1F1F10) - g sin F1F1F1S1F1F1F10) F1F1F1S1F1F1F10t / 2 The negative sign in front of the g term indicates that the velocity in the vertical direction is decreasing due to the Earth's gravity. Therefore, the correct answer is option C: V_o sin F1F1F1S1F1F1F10) - ½ g 2F1F1F1S1F1F1F10t)².

The Figure Represents the Parabolic Trajectory of a Ball Going θ\thetaθ

The Figure Represents the Parabolic Trajectory of a Ball Going $\theta$