 The figure (Figure 1) shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t0=0s corresponds to the moment just after the ball is launched from position x0=0m and y0=0m. Its launch velocity, also called the initial velocity, is vâƒ— 0.

Here is the answer for the question - The figure (Figure 1) shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t0=0s corresponds to the moment just after the ball is launched from position x0=0m and y0=0m. Its launch velocity, also called the initial velocity, is vâƒ— 0.. You'll find the correct answer below

The figure (Figure 1) shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t0=0s corresponds to the moment just after the ball is launched from position x0=0m and y0=0m. Its launch velocity, also called the initial velocity, is vâƒ— 0. Part A How do the speeds v0, v1, and v2 (at times t0, t1, and t2) compare? v0 = v1 = v2 > 0 v0 = v2 > v1 = 0 v0 = v2 > v1 > 0 v0 > v1 > v2 > 0 v0 > v2 > v1 = 0

v0 = v2 > v1 > 0(Here v0 equals v2 by symmetry and both exceed v1. This is because v0 and v2 include vertical speed as well as the constant horizontal speed.)

Reason Explained

v0 = v2 > v1 > 0(Here v0 equals v2 by symmetry and both exceed v1. This is because v0 and v2 include vertical speed as well as the constant horizontal speed.) is correct for The figure (Figure 1) shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t0=0s corresponds to the moment just after the ball is launched from position x0=0m and y0=0m. Its launch velocity, also called the initial velocity, is vâƒ— 0. 