at t = 4 seconds, the ball/s highest point, its speed is briefly equal to:\n0\n12 m/s\n-9.8 m/(s^2)\n40…

at t = 4 seconds, the ball/s highest point, its speed is briefly equal to:\n0\n12 m/s\n-9.8 m/(s^2)\n40 m/s\nquestion 18\n1 pts\na volleyball is bumped upwards. where is its vertical acceleration the greatest? ignore air resistance.

at t = 4 seconds, the ball/s highest point, its speed is briefly equal to:\n0\n12 m/s\n-9.8 m/(s^2)\n40 m/s\nquestion 18\n1 pts\na volleyball is bumped upwards. where is its vertical acceleration the greatest? ignore air resistance.

Answer

Explanation:

Step1: Analyze velocity at highest - point

In projectile motion, at the highest point of the ball's trajectory, the vertical component of velocity is 0. The horizontal component of velocity remains constant throughout the motion (assuming no air - resistance). Given the horizontal velocity $v_x = 12\ m/s$, and vertical velocity $v_y=0$ at the highest point. The speed $v=\sqrt{v_x^{2}+v_y^{2}}$. Since $v_y = 0$ and $v_x=12\ m/s$, the speed at the highest point is equal to the horizontal component of velocity.

Answer:

B. 12 m/s

Explanation for Question 18:

In free - fall motion (ignoring air resistance), the acceleration due to gravity $g$ is constant near the surface of the Earth and is equal to approximately $9.8\ m/s^{2}$ in the downward direction. The vertical acceleration of the volleyball is always $g = 9.8\ m/s^{2}$ (downward) throughout its motion, whether it is moving up, at the highest point, or moving down. So the vertical acceleration is the same everywhere.

Answer for Question 18:

The vertical acceleration is the same everywhere (equal to the acceleration due to gravity, approximately $9.8\ m/s^{2}$ downward).