physics
posted by ruby on .
A person throws a ball upward into the air with an initial velocity of 15 m per second.
 How could you calculate how high the ball goes and how long the ball is in the air before it comes to the thrower's hand?

It travels upwards until the vertical velocity component becomes zero. An easier way to calculate it it to require that the initial kinetic energy become completely potential energy there.
(1/2) M Vo^2 = M g H
where H is the maximum height above the thrower's hand.
H = Vo^2/(2 g) = 11.5 m
The time in the air is twice the time it takes for the velocity upward to become zero. It spends an equal length of time coming back down. Therefore:
T = 2 * (Vo/g) = 3.06 s 
thanx so much! im finally starting to get this!