Question

horizontal velocity is 12 m/s. at t = 4 seconds, the ball/s highest point, its speed is briefly equal to: -9.8 m/(s^2) 12 m/s 0 40 m/s

Explanation:

Step1: Analyze velocity components

In projectile - motion, at the highest point of the trajectory, the vertical component of velocity ($v_y$) is 0. The horizontal component of velocity ($v_x$) remains constant throughout the motion because there is no acceleration in the horizontal direction (assuming no air - resistance).

Step2: Determine the speed at the highest point

The speed $v$ of an object is given by $v=\sqrt{v_x^{2}+v_y^{2}}$. Since $v_y = 0$ at the highest point and $v_x=12$ m/s, then $v = 12$ m/s.

Answer:

12 m/s