# 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.

**By:**Answerout

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

**The Correct Answer is**

**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.)**

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