Aperson throws a ball upward into the air with an initial velocity of 15.0 m/s. calculate (a) how high itgoes, and (b) how long the ball is in the air before it comes back to his hand.

For a) You know that when the ball reaches maximum hight, it has a velocity of 0m/s. You would use the equation ... V is the velocity at the apex and that is 0m/s,  is the initial velocity which is 15.0m/s, 'a' is the acceration and in this case it is gravity which will be -9.8m/s^2 because we decided that up will be positive, and "x" is the distance between your hand and the apex. So you will have 0= 15^2+2(-9.8)x and you should get 11.48m.
For b) you would use the equation x=. In this case x would 11.48m,  is 0m, and t is time.. so you would have 11.48=15t+1/2(-9.8)t^2 and get t= 1.5s ... this is just the time it took to go to its maximum hight so you multiply it by 2 because it takes the same time to get up as it does to come down. You should of gotten 3 seconds. I hope this helped

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