# Consider a diagram of the ball at time t0.(Figure 2) Recall that t0 refers to the instant just after the ball has been launched, so it is still at ground level (x0=y0=0m). However, it is already moving with initial velocity vâƒ— 0, whose magnitude is v0=30.0m/s and direction is Î¸=60.0degrees counterclockwise from the positive x direction.

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Here is the answer for the question - Consider a diagram of the ball at time t0.(Figure 2) Recall that t0 refers to the instant just after the ball has been launched, so it is still at ground level (x0=y0=0m). However, it is already moving with initial velocity vâƒ— 0, whose magnitude is v0=30.0m/s and direction is Î¸=60.0degrees counterclockwise from the positive x direction.. You'll find the correct answer below

### Consider a diagram of the ball at time t0.(Figure 2) Recall that t0 refers to the instant just after the ball has been launched, so it is still at ground level (x0=y0=0m). However, it is already moving with initial velocity vâƒ— 0, whose magnitude is v0=30.0m/s and direction is Î¸=60.0degrees counterclockwise from the positive x direction. Part B What are the values of the intial velocity vector components v0,x and v0,y (both in m/s) as well as the acceleration vector components a0,x and a0,y (both in m/s2)? Here the subscript 0 means "at time t0."

The Correct Answer is

15.0, 26.0, 0, -9.80(Also notice that at time t2, just before the ball lands, its velocity components are v2,x=15m/s (the same as always) and v2,y=âˆ’26.0m/s (the same size but opposite sign from v0,y by symmetry). The acceleration at time t2 will have components (0, -9.80 m/s2), exactly the same as at t0, as required by Rule 2.)

### Reason Explained

15.0, 26.0, 0, -9.80(Also notice that at time t2, just before the ball lands, its velocity components are v2,x=15m/s (the same as always) and v2,y=âˆ’26.0m/s (the same size but opposite sign from v0,y by symmetry). The acceleration at time t2 will have components (0, -9.80 m/s2), exactly the same as at t0, as required by Rule 2.) is correct for Consider a diagram of the ball at time t0.(Figure 2) Recall that t0 refers to the instant just after the ball has been launched, so it is still at ground level (x0=y0=0m). However, it is already moving with initial velocity vâƒ— 0, whose magnitude is v0=30.0m/s and direction is Î¸=60.0degrees counterclockwise from the positive x direction.