Question

A person 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.

Answer 1

For a) You know that when the ball reaches maximum hight, it has a velocity of 0m/s. You would use the equation
`v^(2) = v_(0) ^(2)+2ax`... V is the velocity at the apex and that is 0m/s,
`v_(0)` 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=
`x_(0)+v_(0)t+ (1)/(2)at^(2)`. In this case x would 11.48m,
`x_(0)` 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