Question

a ball is launched from ground at an angle with an initial speed of 10 m/s ($v_x = 6$ m/s, $v_y = 8$ m/s). the ball travels over flat ground. ignore the effects of air resistance. what is the speed of the ball when it reaches the highest point in its trajectory?
a 0.0 m/s
b 6 m/s
c 8 m/s
d 10 m/s

Explanation:

Step1: Analyze vertical and horizontal velocity components at highest point

At the highest point of a projectile's trajectory, the vertical component of velocity \(v_y = 0\) (because the ball stops moving upward and hasn't started moving downward in the vertical direction at that instant), while the horizontal component of velocity \(v_x\) remains constant throughout the motion (since there is no acceleration in the horizontal direction when air - resistance is ignored).
We are given that \(v_x=6\ m/s\)

Step2: Calculate the speed at the highest point

The speed \(v\) of the ball at any point is given by the magnitude of the velocity vector \(v=\sqrt{v_x^{2}+v_y^{2}}\).
Since \(v_y = 0\ m/s\) at the highest point and \(v_x = 6\ m/s\), we substitute these values into the formula: \(v=\sqrt{6^{2}+0^{2}}\)

$$v=\sqrt{36+0}=6\ m/s$$

Answer:

B. \(6\ m/s\)