1. The problem statement, all variables and given/known data A major leaguer hits a baseball so that it leaves the bat at a speed of 30.1 m/s and at an angle of 37.2 degrees above the horizontal. You can ignore air resistance. a) At what two times is the baseball at a height of 10.7 m above the point at which it left the bat? Voy = 18.2 m/s (30.1 sin37.2) Vox= 24.0 m/s (30.1 cos37.2) ax=0 ay= -9.8 m/s^2 yo = 0 xo = 0 vy= -18.2 m/s xy= 24.0 m/s 2. Relevant equations any constant acceleration equations. 3. The attempt at a solution I received Ttotal = 3.71 seconds which I think is right. I set up the equation to find time by doing 0-10.7 = 18.2t + .5(-9.8)t^2 which is -4.9t^2 + 18.2t + 10.7 I did the quadratic equation but received the times -0.516 and 4.23, and added together equal my total time of 3.71, however, I don't understand how there is a negative number. Are these the right times? P.S (this is a 7 part question, so I might need help after this question). Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a). I'm guessing this means use .732 seconds for time. I used the equation: vy=voy +ayt Vy= 18.2+(-9.8)(.732) =11.03 m/s. Is this right?